


-mm\ 



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LIBRARY OF CONGRESS 



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MECHANICAL DRAWING. 



PREPARED FOR THE USE OF THE STUDENTS 

OF THE 

MASSACHUSETTS INSTITUTE OF TECHIOLOGY, 

BOSTON, MA.SS. 



LINUS FAUNCE. 



NINTH EDITION. 



BOSTON: 
Published by the Author, 

MASSACHUSETTS INSTITUTE OF TECHNOLOaY. 

1898. 







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Copyrighted, 1887, by Linus Faunce, 



W. J. ScHOFiELD, Printer, 105 Summer Street, Boston. 



48 6555 

JUL t W2 






CONTENTS. 



Page. 

CHAPTEE I.— Instruments and their Uses. . , o , . 5 
" Definitions, Terms, etc <, . 12 



II. — Geometrical Problems. 



III. — Inking. 
*' Tinting. 



17 

59 
66 



TV. — Projections „ » . 69 

*' Notation » .... 72 

" Projections of Straight Lines. , , . . 73 

*' " " Surfaces. ....,,, 79 

" " " Solids. ........ 85 



V. — Shadows. 



100 



vi. — isometrical drawing. .... o ... 117 
" Oblique Projections . 125 



VIL — Working Drawings. . 

VIII. — EXAIVIPLES o 

3 



. 126 
. 132 



MECHANICAL DRAWING. 



CHAPTER L 

INSTRUMENTS AND THEIR USES. 

1. To do good work good instruments are essential. An 
accomplished draftsman may do fair work with poor instruments, 
but the beginner will find it sufficiently hard to do creditable 
work without being handicapped by poor instruments. It is 
also essential that the instruments should be kept in good order. 
They should be handled carefully, and wiped, before being put 
away, with wash-leather or chamois skin. This is especially 
needful if the hands perspire perceptibly. 

2. Pencilling. Drawings should always be first made in 
pencil, and inked afterwards if desired. The idea of pencilling 
is to locate the lines exactly, and to make them of the required 
length. Accuracy in a drawing can only be obtained by accu- 
racy in the pencil construction. There is a great tendency 
among beginners to overlook this important fact, and to become 
careless in pencilling, thinking they will be able to correct their 
inaccuracies when inking. This is a great mistake, and one to 



6 INSTRUMENTS AND THEIR USES. 

be especially avoided. This accuracy can be obtained in pencil 
only by making very fine, light lines, and to this end hard pen- 
cils, 6 H, should be used, and they should be kept sharp. For 
drawing straight lines the pencil should be sharpened to a flat, 
thin edge, like a wedge. The compass pencils should be sharp- 
ened to a point. A softer pencil, 4 H, sharpened to a point, 
should be used in making letters, figures, etc. 

It should be borne in mind that a 6 H pencil sharpened to a 
chisel point will make a depression in the paper, which can 
never be erased, if much pressure is put on the pencil ; hence, 
press very lightly when using a hard pencil, so as to avoid this 
difficulty. If a drawing is not to be inked, but made for rough 
use in the shop, or where accuracy of construction (in the draw- 
ing) is not essential, or to be traced, a soft pencil would prefer- 
ably be used, the lines being made somewhat thicker, or heavier. 

3. Compasses. In using the compasses the lower part of 
the legs should be kept nearly vertical, so that the needle point 
will make only a small hole in revolving, and both nibs of the 
pen may press equally on the paper. In pencilling it is not so 
essential that the pencil point be kept vertical, but it is well to 
learn to use them in one way, whether pencilling or inking. 

Hold the compasses loosely between the thumb and fore- 
finger only, and do not press the needle point into the paper. 
If it is sharp, as it should be, the weight of the compass will be 
sufficient to keep it in place. While revolving, lean the com- 
pass very slightly in the direction of revolution, and put a little 
pressure on the pencil or pen point. 

In removing the pencil or pen point to change them, be very 
careful to pull them out straight ; do not bend them from side to 
side, in order to get them out more easily, as it would enlarge 
the socket and consequently spoil the instrument for accurate 
work. 



IKSTEUMENTS AND THEIR USES. 7 

In drawing a circle of larger radius than could be drawn with 
the compass in its usual form a lengthening har is used. In this 
case steady the needle point with one hand and describe the 
circle with the other. 

The large compasses are too heavy and clumsy to make small 
circles nicely, hence the bow compasses should be used in mak- 
ing all circles smaller than three-quarters of an inch radius, or 
thereabouts, depending on the stiffness of the spring. Be very 
careful to adjust the needle point to the same length as the pen- 
cil or pen point. In changing the radius of the bow compasses 
or spacers, press the points together, thus removing the pressure 
from the nuts, before turning the nuts in either direction. The 
screw thread will last much longer if this is done. 

4. Dividers or Spacers. These are used to lay off 
distances from scales, or from other parts of a drawing to a line, 
or to divide a line into equal parts. In laying off the same dis- 
tance several times on a line, keep one of the points of the dividers 
on the line all the time, and turn the instrument in an opposite 
direction each time, so that the moving point will pass alter- 
nately to the right and left of the line. Do not make holes in 
the paper in doing this, as it is impossible to ink nicely over 
them; a very slight puncture is sufficient. 

5. T-Square. The T-square should be used with the head 
against the left-hand edge of the drawing board (unless the per- 
son is left-handed), and horizontal lines only should be drawn 
with it. Lines perpendicular to these should not be drawn by 
using the head of the T-square against an adjoining edge of the 
board, as there is no pains taken to make these edges at right 
angles to each other, but they should be drawn by asing the 
triangle in connection with the T-square. 



8 



INSTRUMENTS AND THEIR USES. 



Lines should be drawn with the upper edge only of the 
T-square. 

In case you wish to use the T-square as a guide for the knife in 
cutting paper to size, do not use the upper edge as a guide, but 
turn the T-square over and use the bottom edge. For, unless 
you are very careful, the knife will nick the edge, which would 
render it unfit to draw lines with. 

6. Triangles. To draw lines which shall he parallel to 
another by means of the triangles. Let AB be the given line. 




Place either edge of either triangle so as to coincide exactly 
with the given line. Place the other triangle (or any straight 
edge) against one of the other edges of the first triangle. Then, 
holding the second triangle, or straight edge, securely in this 
position with the left hand, move the first one, still keeping the 
two edges in contact. Any line drawn along the edge which 
originally coincided with the line AB will be parallel to it. 

To draw lines which shall he perpendicular to another hy means 
of the triangles. Let AB be the given line. Place the lono-est 
side of the triangle so as to coincide exactly with the given line. 



INSTRUMENTS AND THEIR USES. 9 

Place the other triangle (or any straight edge) against one of 
the other edges of the first triangle. Then, holding the second 
triangle securely in this position with the left hand, revolve the 
first one so that its third edge is against the second triangle or 
straight edge. Any line drawn along the edge which originally 
coincided with the given line AB will be perpendicular to that 
line. 

The right-hand portion of the figure shows how the two tri- 
angles may be used, in connection with the T-square, to draw 
lines making angles of 15° and 75° with a given line (in this 
case the line which coincides with the edge of the T-square). 
By turning the triangles over, these angles may be drawn in the 
opposite direction. 

Lines making angles of 30°, 45°, and 60° are drav/n directly 
by means of one triangle and T-square or straight edge. 

7. Irregular Curves. To trace an irreojular curve through 
a series of points, use that part of the edge of the curve which 
coincides with the greatest possible number of points (never less 
than three), and draw the curve through these points, then shift 
the curve so as to coincide with other points in the same way, 
letting the instrument run back on a part of the curve already 
drawn, so that a continuous smooth curved line may be formed. 

It requires considerable practice to draw irregular curves by 
means of an instrument, the tendency being to make a series of 
loops, on account of some of the points being covered up. 
There is no better way to put in a curve in pencil than by doing 
it free-hand, provided the hand and eye have been properly 
trained. Of course the curve cannot be inked in free-hand; the 
irregular curve must be used ; but, being no longer confined to 
points, it is not difficult. 



10 INSTRUMENTS AND THEIR USES. 

8. Scales. As it is frequently impossible to make a draw- 
ing on paper the real size of the object, it is customary to reduce 
the actual measurements by means of an instrument called a 
scale, — that is, the drawing may be made ^, ^, ■§■, ^q, etc., size, 
according as the relative size of the object and drawing may 
require. 

If it is desired to make a drawing ^ size, then 3 inches ou 
the drawing will represent one foot on the object. It is fre- 
quently necessary to represent inches and fractions of an inch, 
hence divide the 3 inches into 12 equal parts, and each of these 
parts will represent one inch on the object. If each of the 12 
parts is subdivided into 2, 4, or 8 parts, each part would rep- 
resent respectively ^, ^, or -J of an inch on the object. This 
may be designated, scale, 3 inches equal one foot, or ^ size. 

On the scale, one inch equal one foot, the unit, one inch, is 
divided into 12 parts to represent inches as before. Thus, to 
make a scale of any unit to one foot, it is simply necessary to 
divide that unit into 12 parts to represent inches, subdividing 
these parts, as far as possible, to represent fractions of an inch. 

If the smallest division on a scale represents -J of an inch on 
the object, the scale is said to read to -J of an inch. 

The student will find on his triangular scale ten different 
scales, viz., g^, -J, j\, ^, f, |-, f , 1, 1^, and 3 inches to the foot, 
reading to 2", 1", 1", 1", 1", ^", ■^", ^", J", and -J", respectively. 
On some triangular scales, the scale 3^2" = 1' is divided so as to 
read to 3" instead of to 2". 

The double prime over a number or fraction means inches, 
the single prime indicates feet. 

The scale should never be used as a ruler to draw lines with. 

9. Needle Point. Each student should procure a Jlne 
needle, break off the eye end, and force the broken end into a 



INSTRUMENTS AND THEIR USES. 



11 



small, round piece of soft pine wood. This is to be used in 
pricking off measurements from the scale, marking the exact 
intersection of two lines, etc. Here, as in the case of the needle 
point in the compasses, it should not be forced into the paper; 
the finest puncture possible is sufficient. 



10. Drawing Paper. 

standard sizes, as follows : - 



This paper comes in sheets of 



Cap, . . . 


13 X 17 inches. 


Demy, . . 


15 X 20 " 


Medium, 


17 X 22 " 


Royal, . . 


19 X 21 " 


Super-Royal, 


19x27 " 


Imperial, . 


22 X 30 " 



Elephant, . . 23 x 28 inches. 
Columbia, . . 23 x 31 
Atlas, . . . 26 X 31 
Double Elephant,27 x 10 
Antiquarian, . 31 x 53 
Emperor, . . 18 x 68 

Whatman's paper is considered the best. This paper is either 
hot or cold pressed, the hot pressed being smooth and the cold 
pressed rough. The rough paper is better for tinting work, the 
smooth takes ink lines better than the rough, but erasures show 
much more distinctly on it, hence the cold pressed is better for 
general work. The names of the sizes of the paper given above 
have no reference to quality. There is very little difference 
in the two sides of the paper, but that one which shows the 
maker's name in water lines, when held up to the light, is con- 
sidered the right side. 

11. Thumb Tacks. These are used for fastening the paper 
to the drawing board when it is not necessary to stretch it. 

12. The geometrical problems in the next chapter are not 
given with the view of teaching geometry, but to give the stu- 
dent practice in the accurate use of his instruments. 

In order that the degree of accuracy of the execution of the 
problems may be readily seen, these problems will not be inked. 



12 INSTRUMENTS AND THEIR USES. 

DIRECTIONS FOR MAKING PLATES. 

13. The plates on which these problems are to be drawn 
should be laid out 10" by 14" (and cut this size when finished), 
with a border line one inch from each edge. That portion of 
the plate within the border line is to be divided into 4 equal 
rectangles, in each of which one problem is to be drawn, begin- 
ning with No. 1 in the upper left-hand corner rectangle, No. 2 
in the upper right, etc. 

The problems should be drawn as large as the size of the 
rectangles will permit, as the larger they are the easier it is to 
obtain the accurate results which will be insisted upon. 

The number of the plate is to be printed in Arabic numerals 
in the upper right-hand corner of the plate, about one-eighth of 
an inch above the border line, and the student's name in the 
lower right-hand corner, about one-eighth of an inch below the 
border line. 

Prob. 1, Prob. 2, etc., as the case may be, is to be printed 
in the upper right-hand corner of each rectangle. The first 
letter of Prob. and the initial letters of the name should be 
made ^q of an inch high, and the small letters -J of an inch 
high ; the numerals should also be yV' high. 

The letters and figures should be made like those on page 16, 
and they should be made as nicely as possible. 

The small letters should be made with the same style of let- 
ter as the initials. 

DEFINITIONS, TERMS, ETC. 

14. In Geometry, space is described in terms of its three 
dimensions, — length, breadth, and thickness. 

A point has simply position without any dimension. 

A Uneh'ds only one dimension, viz., length. 

A surface has two dimensions, — length and breadth. 



INSTRUMENTS AND THEIR USES. 13 

A solid has three dimensions, — length, breadth, and thickness. 

Lines are of several kinds, — straight or right lines, curved 
lines, and broken lines. 

Parallel lines are everywhere equally distant from each other. 

An angle is the difference in direction of two lines. The 
point of meeting of these lines or sides, produced if necessary, 
is the vertex of the angle. 

A right angle is one where its two sides make an angle of 90° 
with each other. 

An acute angle is less than a right angle. 

An obtuse angle is more than a right angle. 

A quadrant is an arc of 90°. 

A plane Jigure is a plane bounded on all sides by lines. If 
the lines are straight, the space which they contain is called a 
rectilinear Jigure or polygon, and the sum of the bounding lines 
is the perimeter of the polygon. 

Polygons are named according to the number of their sides, 
thus : — 

A triangle is a plane figure of three sides. 

A quadrilateral is a plane figure oifour sides. 

A. pentagon is a plane figure oijive sides. 

A hexagon is a plane figure of six sides. 

A heptagon is a plane figure of seiien sides. 

An octagon is a plane figure of eight sides. 

A nonagon is a plane figure of nine sides. 

A decagon is a plane figure of ten sides. 

An undecagon is a plane figure of eleven sides. 

A dodecagon is a plane figure of twelve sides. 

A circle is a plane figure of an infinite number of sides, or a 
plane figure bounded by a curved line everywhere equi-distant 
from its centre. 

An equilateral triangle has all its sides and angles equal. 

An isosceles triangle has two sides and two angles equal. 



14 INSTRUMENTS AND THEIR USES. 

A scalene triangle has all its sides and angles unequal. 

Quadrilaterals have a variety of names according to the 
relative positions of their sides, thus : — 

A trapezium is a quadrilateral having no two sides parallel. 

A trapezoid is a quadrilateral having two only of its sides 
parallel. 

A parallelogram is a quadrilateral having its opposite sides 
parallel. Parallelograms are named as follows : — 

A square is a parallelogram having all of its sides equal and 
its angles right angles. 

A rectangle is a parallelogram having its opposite sides equal 
and its angles right angles. 

A rhombus or rhomb is a parallelogram having all its sides 
equal, but the angles are not right angles. 

A rhomboid is a parallelogram having its opposite sides equal, 
but its angles not right angles. 

A polygon which has all its vertices on the circumference of 
a circle is said to be inscribed in the circle. The circle is cir- 
cumscribed about the polygon. 

A polygon which has all its sides tangent to a circle is said 
to be circumscribed about the circle. The circle is inscribed in 
the polygon. 

A diameter is any line drawn through the centre of a figure, 
and terminated by the opposite boundaries. 

The long diameter of a polygon is the diameter of its circum- 
scribed circle. This is also called a diagonal. 

The short diameter of a polygon is the diameter of its in- 
scribed circle. 

All polygons are supposed to be regular unless otherwise stated. 

A polyhedron is a solid bounded entirely by planes. 

The edges of a polyhedron are the lines of intersection of its 
bounding planes. 



INSTRUMENTS AND THEIR USES. 15 

The sides or faces of a polyhedron are the plane figures com- 
posing its surface. 

A polyhedron is said to be regular when its faces are equal 
and regular polygons, and each adjacent pair include the same 
angle. There are only five regular polyhedrons, viz. : — 

The tetrahedron, bounded by four equal equilateral triangles. 
The hexahedron or cube, bounded by six equal squares. The 
octahedron, bounded by eight equal equilateral triangles. The 
dodecahedron, bounded by twelve equal pentagons. The icosa- 
hedron, bounded by twenty equal equilateral triangles. 

A prism is a polyhedron having two of its faces, called it& 
ends or bases, parallel, and the rest parallelograms. 

A parallelopiped is a prism whose bases are parallelograms. 

A pyramid is a polyhedron having a polygon for its base, 
and for its sides it has triangles which have a common vertex, 
and the sides of the polygon for their bases. 

The common vertex of the triangles is called the vertex, or 
apex, of the pyramid. 

The axis of a prism is a straight line joining the centres of 
its ends ; and the axis of a pyramid is the straight line from its 
vertex to the centre of its base. 

A right prism or pyramid has its axis at right angles to 
its base. 

Prisms and pyramids are named from the form of iheir bases, 
as triangular, square, pentagonal, hexagonal, etc. 

A cylinder is similar to a prism except that its bases are 
circles. 

A cone is similar to a pyramid except that its base is a circle. 

All prisms, pyramids, cylinders, and cones are supposed to be 
nght unless otherwise stated. 



16 INSTRUMENTS AND THEIR USES. 

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CHAPTER 11. 



GEOMETRICAL PROBLEMS. 



Prob. 1. To hisect a straight line AB, or arc of a circle AFB. 
With A and B as centres and any radius 
greater than one half AB draw arcs intersect- 
ing in C and D. Join CD. CD is perpen- 




^ dicular to AB, and E and F are the middle 
points required. 

Note. To draic a perpendicular to a line 
at any point in it, as E in AB. Lay off equal distances EA 
and EB on each side of E, and proceed as above. 



Prob. 2. From a point C outside a straight line AB to draw 
a perpendicular to the line, 

c With C as a centre and any convenient 

radius cut AB in the points A and B. With 

B,^ A and B as centres and any radius draw arcs 

.--'' intersecting in D. Join C and D, and CD is 

the perpendicular required. 

17 



18 GEOMETEICAL PEOBLEMS. 

pROB. 3. To draw a 'perpendicular to a line K^ from a point 
C nearly or quite over its end. 

_,Jjl. Draw a line from C to meet AB in any 

point B. Bisect BC in D by Prob. 1. With 

'v / D as a centre and radius DC draw the arc 

/' -^ CAB meeting AB in A. Join A and C, 

"^ ,B and AC is the perpendicular required. 

'» "^-- '' Note. If a perpendicular be required at 

A, take any point D as a centre and radius DA and draw an 
arc CAB. Through B and D draw a line to meet the arc in C. 
Join A and C, and AC is the perpendicular required. 



Prob. 4. To draw a perpendicular to a line A^ from a point 
A at or near its end. 

g— With A as a centre and any radius draw 

the arc CD. With centre D and same radius 
\g^ cut CD in C. With C as a centre and same 

'"n radius draw an arc over A, and draw a line 

V, „ throuo^h D and C, producino^ it to meet this 
arc in E. Join A and E, and AE is the per- 



pendicular required. 



Prob. 5. A second method. 

With centre A and any radius draw an arc 

CDE. With centre C and same radius cut this 

\ arc in D. With centre D and same radius draw 

r-^<[ arc EF. With centre E and same radius draw 

\ arc intersectino: EF in F. Join A and F, and 

S — S AF is the perpendicular required. 



-".E 




GEOMETRICAL PROBLEMS. 19 

Prob. 6. To construct an angle equal to a given angle CAB. 

9^" Draw any line as DE, and take any point in 

y^ \ it as D. With A as a centre and any radius 

4 ^ cut the sides of the given angle in B and C. 

With centre D and same radius draw arc EF. 
With BC as a radius and E as a centre cut arc 
• EF in F. Join D and F, and FDE is the 
angle required. 

Note. For accuracy of construction in drawing this prob- 
lem the longer the radius AB is the better. 

Prob. 7. Through a given point C to draw a line ^parallel to 
a given line AB. 
v^ c_ From C as a centre and any radius 

\ ^'' \ draw the arc AD ; and from A as a centre, 

\ y \ with the same radius, draw the arc BC. 

—^ ^ With BC as a radius, and A as a centre, 

draw an arc cutting the arc AD in D. Join D and C, which 
is the line required. 

Prob. 8. To draw a line -parallel to a given line AB at a 

given distance CD from it. 

,-. ^ -.^^ ^,-. >f . -^ From any two points A and B on the line 

' '1 as centres and with CD as a radius draw arcs 

I y ] ^^ E and F. At A and B erect perpendiculars 

c, ^D to meet the arcs in E and F. Draw a line 

through E and F, which is the line required. 

Prob. 9. To bisect a given angle BAC. 

^ With A as a centre and any radius draw the 

^y\ arc BC, cutting the sides of the angle in B and 

^y^ \ ^^' C. With centres B and C and any radius 

^""■^-.^^^ /' ' draw arcs intersecting in D. Draw AD, which 

c^\^ will bisect the angle. 



20 



GEOMETRICAL PEOBLEMS. 



Prob. 10. To divide a given line AB into any number of 
equal parts. (In this case 6.) 

^^..-f'" From A draw an indefinite line Al, 2 • • 5 

^^'f ! J\ at any angle with AB. At B draw Ba " ' e, 

h'!'; / / / J B making the angle ABe equal to the angle 

BA5. With any distance as a unit, lay off on 

the lines from A and B as many equal spaces 

as the number of parts required less one. 

Join le, 2c?, 3c, etc., and where these lines intersect AB are the 

points of division required. 



'fe 



.-'6 



Prob. 11. Another method. To divide a line A.^ into (^'a,j) 
jive equal parts. 

Draw Al 5 at any angle to AB, and 

lay off on it five equal spaces, using any con- 
venient unit. Join 5B, and through the 
points 1, 2, 3, 4 draw lines parallel to Bo, 

meeting AB in points a, h, c, and d, which are the points of 

division required. 




Prob. 12. To divide a line AC into the same proportional 
parts as a given divided line AB. 

Draw from a point A the lines AC and 
AB, making any angle with each other. Join 
B and C. Through the points 1, 2, and 3 on 
AB draw lines parallel to BC, meeting AC 
^ in points a, h, and c, the required points of 
division. 




GEOMETRICAL PROBLEMS . 



21 



Second method. 

Let M be the line to be divided into parts 
proportional to the parts Al, 12, 23, and 3B 
of the line AB. Draw CD equal in length to 
M and parallel to AB by Prob. 8. Draw lines 
throuojh AC and BD to meet in E. Throui^h 
E draw lines to 1, 2, and 3, cutting CD in a, 
5, and c, the required points of division. CD can be placed at 
any distance from AB, provided the lines AC and BD intersect 
within the limits of the paper. 

Note. If the parts on AB are equal, the parts on CD will 
be equal. 



Prob. 


13. 


^ 


M:. 






' 1 \ * 
'/ 1 \ 


'\^ 


n 


' ' "^ 


\ 




^' ^^. 


3 


B 



Prob. 14. To draw on a given line AB as an hypothenuse a 
right triangle with its sides having the proportion 0/3,4, and 5. 
Divide AB by Prob. 11 into 5 equal parts. 
With centre B and a radius equal to three of 
the parts draw an arc, and with centre A and 
a radius equal to four of the parts cut this 
arc in C. Join AC and BC, and ACB is 
the required right triangle. 




Prob 




To trisect a right angle CAB. 

With centre A and any radius draw the arc 
of the quadrant, cutting the sides in C and B. 
With centres C and B and the same radius cut 
the arc in points 1 and 2. Join Al and A2. 

Note. The angle lAB is an angle of 60°, 



the construction of which is apparent. 



22 



GEOMETEICAL PROBLEMS. 



Prob. 16. 

and N. 



- t5 



To find a mean proportional between two lines M 

Upon an indefinite line AB lay off AC equal 
to M, and CB equal to N. Bisect AB in E 
I by Prob. 1, and with centre E and radius EB 
■* draw a semicircle. At C draw CD perpen- 
dicular to AB (Prob. 5). CD is the mean 
proportional required. 



Prob. 17. To find a fourth proportional to 
N, and P. 



lines M, 



portional required. 



Draw AB equal to M, and at any conve- 
nient angle draw AC equal to N. Upon 
AB produced make BD equal to P. Join 
BC, and draw DE parallel to BC, to meet 
AC produced in E. CE is the fourth pro- 



Prob. 18. 
80^ 




At a point A in a line AC to make an angle of 



With any point B as a centre and radius 
AB draw a semicircle ADC. With centre C 
and same radius cut this arc in D. Draw 
AD» DAC = 30°. 



Prob. 19. 




Having given the sides AB, M, and 'N of a tri- 
angle to construct the figure. 

With centre A and radius M draw an arc. 
With B as a centre and radius N draw an arc 
to cut the first arc at C. Join AC and BG. 
ABC is the triangle required. 



GEOMETRICAL PEOBLEMS. 



2B 








n a given 



AB to construct a square. 



Draw BD at right augles to AB and equal 
/\ to AB (Prob. o). With A and D as centres 
and radius AB draw arcs intersecting in C. 
Join AC and CD. 



Prob. 21. To construct a rectangle of given sides AB and M, 

^ D_^ At B draw BD perpendicular to AB by 

>; Prob. 5, and equal to M. With A as a cen- 
x---^tre and radius equal to M draw an arc, and 

J i-^—^^ from D as a centre and a radius equal to AB 

cut this arc in C. Join AC and CD. 



Prob. 22. On a given diagonal AB to construct a rhombus 
of given side AC. 

With centres A and B and radius AC draw 
arcs intersecting in C and D. Join AC, AD,- 
i BC, and BD. 




Prob. 23. 

'pentagon. 




.,-'f^- 



On a given hase AB to construct a regular 



With centres A and B and radius AB 
draw circles intersecting in 1 and 2. Join 
1 and 2. With centre 2 and same radius 
draw the circle 3A5B4, giving points 
3 and 4. Produce 35 to C and 45 to 
E. With centres C and E and radius 
AB draw arcs intersecting in D. Draw 
BCDEA. 
Note. This is an approximate method. 



24 



GEOMETEICAL PEOBLEMS. 



Prob. 24. To construct a regular hexagon of 'given side ABo 

^p With A and B as centres and radius AB draw 

\^ arcs intersecting in O. With centre O and 

iQ radius AB draw a circle, and lay off BC, CD, 

etc., each equal to AB. Join the points B, C, 

D, E, F, and A. 

Note. The radius of any circle goes around the circumfer- 
ence as a chord six times exactly. 




Prob. 25. On a given base AB to construct a regular poly- 
gon of any numher of sides (in this case 7). 

With centre A and radius AB draw a 
semicircle and divide it at points 1, 2, 3, 4, 
etc., into as many equal parts as there are 
sides in the required polygon. Draw a line 
from the second point of division 2 to A. 

2A is one side of the required polygon. 

Bisect AB and A2 by perpendiculars, by 

in D. W^ith D as a centre and radius DA 

Apply AB as a chord to the circle 



d-z.':-'. 




Prob. 1, meeting 

draw the circle BA2, etc. 

as many times as there are sides in the polygon. 



Prob. 26. On a given line AB to construct a regular poly- 
gon of any number of sides. An approximate method. 

Bisect AB, and produce the bisecting 
line indefinitely. With centre A and 
radius AB draw the arc BC, cutting the 
bisecting line in C. Divide the arc BC 
into six equal parts, in points 1, 2, 3, etc. 
To construct a pentagon. With centre 
C and radius Cl draw an arc, cutting 
the bisecting line in a point below C, 




GEOMETRICAL PROBLEMS. 



25 



which is the centre of the circle circumscribing the pentagon. If 
the polygon is to have more sides than six, set up from C on the 
line Cah as many parts of the arc CB as added to six make 
the number of sides of the required polygon ; thus, for a seven- 
sided polygon set up one division as Qa ; for an eight-sided set 
up two divisions as Qah, and so on. a, b, c, d, etc., are the cen- 
tres for the circumscribing circles of the polygons, each side of 
which is equal to AB. 



Prob. 27. 
ffon. 



On a given base AB to construct a regular octa- 



At A and B erect perpendiculars by 
Prob. 5 to AB, and bisect the exterior right 
angles, making the bisectors AC and BD 
each equal to AB. Draw CD, cutting the 
perpendiculars in E and F. Lay off EF 
from E to Gr and from F to H. Draw 
an indefinite line through GH. Make GK, GL, HN, and HM 
each equal to CE or FD. Connect C, K, L, M, N, and D. 

Note. Tliis may be done by Prob. 25. Check the con- 
struction by seeing if AN, CM, BD, and KL are parallel. 
CA, KB, etc. should be parallel ; so also AD, CN, and KM. 





To construct a regular octagon within a square 



Draw the diagonals AC and BD. With A, 
B, C, and D as centres and the half of the 
diagonal of the square as a radius draw the 
arcs EF, GH, KL, and MN. Join the points 
MG, FL, HN, and KE. 



26 



GEOMETRICAL PROBLEMS. 



Prob. 29. The altitude AB of an equilateral triangle being 
iven, to construct the triangle. 

Draw CD and EF, both perpendicular to 
AB. With A as a centre and any radius 
describe the semicircle CGHD. With C and 
D as centres and the same radius draw arcs 
cutting the semicircle in Gr and H. Draw 
AGE and AHF. 





> 


k 


-- 


-- 


\ 


// 




\ 


> 




// 




\\ 


/ 


; 




- 




\ 


E 




3 




F 




Prob. 30. Given the base AB of an isosceles triangle, and 
the angle at the vertex M, to draw the triangle. 

With centre B and any radius draw a 
semicircle cutting the base produced at C. 
Make the angle DBG equal to M by Prob. 
6. Bisect ABD by Prob. 9. Make the 
angle FAB equal to EBA, and produce 
.- AF and BE to meet in G. 

Prob. 3L Within a regular hexagon ABCDEF to inscribe 
a square. 

Draw a diagonal AD. Bisect AD by 
a perpendicular 213 (Prob. 1). Bisect by 
Prob. 9 the right angles 21A and 21D, and 
produce the bisectors to meet the sides of the 
hexagon in points G, H, L, and K. Join G, 
H, L, and K. 

Prob. 32. On a given diagonal AB to construct a square. 

Bisect AB in O (Prob. 1) ; and with centre 
O and radius OA draw a circle to cut the 
^B bisecting line in C and D. Draw AC, CB, 
BD, and DA. 





GEOMETEICAL PROBLEMS. 



27 




Within a given triangle ABC to inscribe a square. 

Draw AD perpendicular to and equal to AB. 
From C draw CE perpendicular to AB (Prob. 2). 
Draw DE cutting AC in F. From F draw FK 
perpendicular to AB. Make KH equal to KF, 
and from H with radius HK cut BC in G. Join 



FG and GH. 



Prob. 34. About a given circle FEGD to circumscribe an 
equilateral triangle. 

Draw the diameter DE. With centre E 
and radius of the given circle draw the cir- 
cle AFG. Prolong DE to A. With centres 
D, F, and G and radius DG draw arcs inter- 
secting at B and C. Join the points A, B, 
and C, and ABC is the required triangle. 




To circumscribe a circle about a triangle ABC. 
By Prob. 1 bisect two of the sides AB 
and BC by perpendiculars meeting in O. 
With centre and radius OA draw the 
circle. 

Note 1. If any three points are given 
not in the same straight line, a circle is 
passed through them by joining the points 
and proceeding as above. 

Note 2. If any circle is given, its centre is found by taking 
any three points in its circumference, and proceeding as above. 




28 



GEOMETRICAL PROBLEMS. 



Prob. 36. To inscribe a regular pentagon within a circle. 

Draw any diameter AB, and a radius. CD 

perpendicular to it. Bisect BC in E. With 

centre E and radius ED draw the arc DF. 

JB With centre D and radius DF draw the arc 

FG. DG is the side of the required pen= 




Prob. 37. To inscribe a regular hexagon within a circle. 

Draw the diameter AB, and with centres A 
and B and the radius of the given circle 
draw arcs COD and EOF, cutting the circum- 
ference in C, D, E, and F. Join AD, DF, 
FB, etc. 

Note. Joining points A, E, and F gives 
an inscribed equilateral triangle. 




Prob. 38. To construct a regular 'polygon of any number of 
sides^ the circumscribing circle being given. 

Draw any diameter AB and divide it 
into as many equal parts as there are sides 
in the required polygon (in this case 5). 
With A and B as centres and radius AB 
draw arcs intersecting in C. Draw a line 
from C through the second point of di- 
vision of AB to meet the circumference 
in D. AD is one side of the required 
polygon. Lay off AD as a chord as many 
times as there are sides in the required 
polygon (in this case 5). 

Note. This is an approximate method. 




V^?' 



GEOMETBICAL PROBLEMS. 



29 




Prob. 39. To circumscribe a regular hexagon about a circle. 
Draw the diameters AB and CD perpen- 
dicular to each other. Divide each quadrant 
into thirds, by Prob. 15, at points E, F, etc. 
Join B and E, cutting CD in G. With G 
as a centre and radius GE draw arc ED, 
cutting COD in ^ D. With centre O and 
radius OD cut the diameters produced in points H, K, C, L, and 
M. Join points H, D, M, L, C, and K. . 

Note. Practically the 60° triangle placed on a T-square 
whose blade is parallel to COD will give HD, DM, etc., by- 
making it tangent to the circle at E, F, etc. KH and LM are 
drawn by the T-square. 

Prob. 40. To circumscril 
about a circle. 



a square, also a regular octagon, 




Prob. 41. 

without it. 



Draw the diameters AB and CD at right 
angles to each other. With centres A, B, C, 
and D and radius OA describe arcs intersect- 
ing in points E, F, G, and H. These points con- 
nected give a square about the circle. Inscribe 
an octagon in the square by Prob. 28. 

To draw a tangent to a circle B from a point A 



Draw AB and bisect it in D. With 
centre D and radius DA draw a semi- 
circle cutting the given circle in C and E. 
Join AC. By joining AE a second tan- 
gent is found, equal to AC. 
Note. To draw a tangent to a circle from a point C on the 
circumference. Join BC, and at C draw AC perpendicular to 
BC. For the tangent is perpendicular to the radius at the point 
of tangency. 




80 



GEOMETRICAL PROBLEMS. 




Prob. 42. To draw a tangent to the arc of a circle when the 
centre is not accessible. 

Let C be the point upon the given arc, 
AB, at which the tangent is to be drawn. 
Lay off equal distances upon the arc from 
C to A and B. Join A and B. Through C draw a line par- 
allel to AB by Prob. 7. , 

Prob. 43. To draw a tangent at a given point A on a circle 
when the preceding method is not applicable. 

From A draw any chord AB. Bisect AB in 
C, and the arc ADB in D by Prob. 1. With 
A as a centre and a radius AD draw the arc 
EF. With D as a centre and radius DF draw 
an arc cutting EF at E. Join AE. 

Prob. 44. To draw a tangent to two given circles, A and B. 

Through A and B draw a line. 
Make DH equal to the radius BF. 
Draw the circle A-HC, and from B 
draw the tangent BC by Prob. 41. 
Draw AC and produce it to E. Make 
the angle FBK equal to CAH. Join EF. 

Prob. 45. To draw a tangent to two given circles which 
shall pass between them. 

Join A and B, and draw AD and BC 
perpendicular to AB. Draw DC, cut- 
ting AB in E. Draw a tangent from 
E by Prob. 41 to the given circles. 
Join the tangent points F and G-. FG 
is the required tangent. 





GEOMETRICAL PROBLEMS. 



31 



Prob. 46. To draw a circle tangent to a given line AB at 
a given point B in it, which shall also pass through a fixed 
'point C without the line. 

Draw BD perpendicular to AB, at the 
point B. Join CB, and draw a perpen- 
dicular to it at its middle point, Prob. 1. 
The intersection of this perpendicular and 
BD gives D the centre required. 





Prob. 47. To draw a circle of a given radius AB, which 
shall he tangent to a given circle C, and also to a straight line 
DE. 

Draw GH parallel to DE, by Prob. 
8, at the distance AB. With a radius 
CM, equal to the radius of circle C 
plus AB, draw an arc to meet GH in 
O. With the centre and radius AB 
draw the required circle. 
Note. If two circles are tangent, the straight line connect- 
ing the centres passes through the point of tangency. This 
point it is very important to locate precisely in all cases of tan- 
gency. 

Prob. 48. To draw a circle tangent to a given circle D, and 
also tangent to a given line AB, at a given point B on the line. 

Draw BC perpendicular to AB, and 

make BC equal to the radius of circle D. 

Join DC, and at its middle point draw 

- a perpendicular to meet BC in E, the 

required centre. 

'^ '^ic Note. By laying off BC above the 

line AB, and proceeding as above, another circle is found 

below AB, 




32 



GEOMETRICAL PROBLEMS. 




Prob. 49. To draw a circle tangent to a given circle K at a 
given point E on the circle^ and to a given line BC. 

Draw AE and produce it. At E 

draw a perpendicular to AE, meeting 

, BC in B. Draw the bisector of the 

angle EBC, meeting AE produced 

in D. DE is the radius required. 

Note. By bisecting EBM another 
circle, whose centre is G, is found, 
enclosing the circle AE. 

Prob. 50. To draw a circle of given radius R tangent to 
two given lines, AB and CD. 

Draw the lines CE and AE 
parallel respectively to AB and 
CD, at the given distance R from 
them by Prob. 8, meeting in E. 
With E as a centre and radius 
equal to R describe a circle, which 
will be tangent to the given lines. 




Prob. 51. To draw any number of circles tangent to each 
other, and also to two given li?ies AB and AD. 

Bisect BAD by AC, Prob. 9. Let 
one of the circles be BED, drawn by 
H ^ '^c taking C as a centre, and a radius 
equal to the perpendicular CD from 
C to AD. At E draw EF perpendic- 
ular to AC. With centre F and ra- 
dius FE draw the arc DEG, cutting AD in G. At G make 
GH perpendicular to AD. H is a centre required. Repeat 
the process. 



GEOMETfllCAL PROBLEMS. 



33 



Prob. 52. To draw a circle through a given point D and 
tangent to two given lines AB and AC. 

Draw AD. Bisect BAG by AE. 
From any point K, on AE as a 
centre, draw a circle tangent to AB 
and AC, and cutting AD in H. 
Draw HK. At D draw DE, mak- 
ing the angle ADE equal to AHK. 
E is the centre of the required circle. 




Prob. 53. To draw a circle of a given radius R tangent to 
two given circles A and B. 

Through A and B draw indefinite 
lines, and make DE and EG each 
equal to R. With A and B as centres, 
and radii AE and BG, draw arcs cut- 
ting each other at C, which will be the 
centre of the required circle. 

Note. These arcs will intersect in 
a second centre on the other side of the given circles. 




Prob. 54. To draw a circle through a point C, and tangent 
to a given circle A, at a point B in its circumference. 



Draw AB and produce it indefinitely. 
Join BC and bisect it by a perpendicular 
meeting AB produced in D, which will 




be the centre of the required circle. 



34 



GEOMETRICAL PROBLEMS. 



Prob. 55. Given two parallel straight lines AB and CD, to 
draw arcs of circles tangent to them at B and C, and passing 
through E, which is anywhere on the line BC. 

.c fi_ At B and C erect perpendiculars. 

Bisect BE and CE by perpendiculars 
(Prob. 1), meeting the perpendiculars 
from B and C in F and H, the re- 
quired centres. Draw the arcs BE 

This is called a reversed curve. 




cy-v 



Prob. 56. To draw a circle tangent to two given circles A 
and ^ at a given point C in one of them. 

Draw a line indefinitely through 
A and C. Make CD and CE each 
equal to the radius of circle B. 
Join BE and BD. Bisect BE and 
BD by perpendiculars -meeting AC 
produced in F and G, the centres 
of the required circles. 
Note. The tangent points should 
be determined accurately. 




Prob. 57. 




L, another centre required. 



Same as Prob. 56. A second method. 

Draw a line through A and C 
indefinitely. Through B draw DE 
parallel to CA, cutting circle B in 
D and E. Draw CD to F and CE, 
cuttino; circle B in H. Through B 
and H draw BHK, cutting CA in 
K, one centre required. Through 
B and F draw FBL, cutting CA in 



GEOMETRICAL PROBLEMS. 



35 



Prob. 58. Same as Proh. h^. Prob. 59. Same as Proh. bQ, 

Method of Proh, 57. 





pROBS. 60 and 61. Same as Proh. b^. Method of Proh. 57. 





Prob. 62. 




To inscrihe a circle within a triangle ABC. 

Bisect two of the angles of the triangle, 

,^- say A and B, by Prob. 9. The bisectors 

meet in D, the centre of the required circle. 

A perpendicular from D to either side is the 

^ required radius. 



GEOMETRICAL PROBLEMS. 



Prob. 63. Within an equilateral triangle ABC to inscribe 

three equal circles, each tangent to two others, and to two sides 

of the triangle. 

Draw the bisectors of the angles A, B, 

and C, cutting the sides in D, E, and F. 

With centres D, E, and F, and radius DF, 

draw arcs cutting the bisectors in H, L, 

and K, the required centres. 




Prob. 64. Within an equilateral triangle to draw three egual 
circles, each tangent to two others and to one side of the triangle. 

Bisect the angles A, B, and C. Bi- 
sect the angle DAB by AG. G is the 
centre of one of the required circles. 
With the centre of the triangle as a 
centre, and radius OG, draw a circle 
cutting AD in H and BE in K. With 
centres G, H, and K, and radius GF, 
draw the circles G, H, and K. 




Prob. 65. Within an equilateral triangle ABC to draw six 

equal circles which shall he tangent to each other and to the sides 

of the triangle. 

Inscribe the three circles K, H, and 

G by Prob. 64. Draw LGM parallel 

to AB, MHN parallel to BC, and 

LKN parallel to AC. The points L, 

M, and N are the centres of the other 

three circles. 




GEOMETKICAL PROBLEMS. 



37 



Prob. 66. Within a square ABCD to draio four equal cir- 
cles each tangent to two others and to two sides of the square. 

Draw the diagonals AC and BD, and 
the diameters EF and GH. Draw EH, 
HF, FG, and GE, cutting the diagonals 
in the points M, N, 0, and P, which will 
be the centres of the required circles. 
The radius OR is found by joining O 
and P. 




Prob. 67. Within a given square ABCD to draw four equal 
circles each tangent to two others and to one side of the square, 

^ — -^ f ' V ^ Draw the diagonals AC and BD, and 

the diameters EF and GH. Bisect the 
angle GAB by AK, cutting EF in K. 
H With radius OK and centre O draw a cir- 
cle cutting the diameters in the points L, 
M, N, and K, the required centres. 




Prob. 68. Within a square ABCD to draw four equal semi- 
circles, each tangent to two sides of the square, and their diame- 
ters forming a square. 

^ — ^ Draw the diagonals and diameters. Bi- 
sect FC and BH in L and K. Draw LK, 
cutting GH in M. Set off SM from S on 
;-i^^H the diameters to N, 0, and P. Join the 
points M, N, O, and P. The intersections 
of these lines with the diagonals give the 
required centres, 1, 2, 3, and 4. 




38 



GEOMETRICAL PEOBLEMS. 



Prob. 69. Within a square ABCD to draw four equal semi' 
circles, each tangent to one side of the square and their diameters 
forming a square. 

Draw the diagonals and diameters. Draw 
EH, HF, FG, and GE, cutting the diago- 
nals in the points K, L, M, and N. The 
H lines joining these points cut the dianaeters 
in points 1, 2, 3, and 4, the required cen- 




--V-J 



-iJ-.-^ 



_^2„-I, 



''¥■ 



tres. 



Prob. 70. To draw within a given circle ABC three equal 
circles tangent to each other and the given circle. 

~ Divide the circle into six equal parts 

by diameters AE, DC, etc. (Prob. 24). 
Produce any diameter, as AE to G, mak- 
ing EG equal to the radius of the given 
circle. Join CG. Bisect the angle OGC 
"~----^ by GH, intersecting OC in H. With 
centre and radius OH draw the circle HKLM. K, L, and 
M are the centres of the required circles. 




Prob. 71. To draw within a given circle ACDB/oMr equal 
circles which shall he tangent to each other and the given circle. 
Draw the diameters AB and CD at right 
angles, and complete the square FBED. 
Draw the diagonal FE and bisect the angle 
FED by EG ; FD and EG meet in G. 
With centre F and radius FG draw a cir- 
cle cutting the diameters in G, H, K, and 
r^ L, the required centres. 





GEOMETEICAL PROBLEMS. 89 

Prob. 72. Within a given circle AFD . . . . C to draw six 
circles tangent to each other and the given circles. 

Draw the diameters AB, CD, and EF, di- 
viding the circle into six equal parts (Prob. 
24, Note). Divide any radius as OB into 
three equal parts (Prob. 11), at points F and 
G. With centre and radius OF draw a cir- 
_% cle giving the centres F, 1, 2, 3, 4, 5 required. 
A circle of the radius FB may be drawn 
S B T from centre tangent to the six circles. 
Note. The above is a special method. The general method, 
to draw any number of equal circles in a given circle, tangent to 
each other and the given circle, is to divide the circle by diame- 
ters into twice as many equal parts as circles required. From 
B, the extremity of any one of these diameters, draw a tangent 
ST, Prob. 41, Note. Produce the diameters on each side of 
AB to meet the tangent in S and T. Bisect the angle T. The 
bisector meets OB in F, the centre of one of the required cir- 
cles. Or F may be obtained by making TM equal to TB, and 
at M drawing a perpendicular to OT, meeting OB in F. With 
centre O and radius OF draw a circle cutting the diameters in 
points 1, 2, 3, 4 etc., the centres required. 

Prob. 73. About a given circle to circumscribe any number 
of equal circles tangent to each other and the given circle. 

^"■'"^ --^ Divide the circumference of the given 

circle by diameters into twice as many parts 
as circles required. From the extremity B 
of any diameter draw a tangent SBT (Prob. 
41, Note), and produce the diameters on 
each side of OB to meet SBT in S and T. 
Produce OT, making TN equal to TB. 
Make NR perpendicular to TN, meeting 




40 GEOMETRICAL PROBLEMS. 

OB produced in R, the centre of one of the required circles. 
The other centres are at the intersection of a circumference 
drawn with radius OR and centre O, and every other diameter 
produced. 

Prob. 74. Within a given circle AD . . .V» to draw any num- 
her of equal semicircles tangent to the given circle^ and their diam- 
eters forming a regular 'polygon. 

y^ sZ--^^^^^^ ^"^ ^^ given circle let OA and OB 

/ ,^'' ; I r~""N ^\ t)e two radii at right angles. Divide the 
f v\ / 1 '\ -V \ given circumference, commencing at B, 
[1 ; _yl^j^:-i_'-_'j^_ j|_ Jj into twice as many parts as semicircles 
1\ \ z'^'" I ""^ '\ lyj required, and draw diameters to the 
\ Bt, [r y^-J poiiits of division. Join BA. BA cuts 

\V''"--._j.,--7'y// the first diameter to one side of OB at 
^^^=5s=a..jLll«i*^^^^ G. G is one end of two adjacent diam- 

eters required. Lay off OG from O on every other diameter in 
points H, K, etc. Join HK, KG, etc. 1, 2, 3, etc., are the 
centres of the required semicircles. 

Prob. 75. To divide a circle into any number of parts which 

shall be equal in area. 

Let the number of parts be four. Divide 
a diameter into twice as many equal parts 
as areas required, in this case eight, by 
points 1, 2, 3, 4, etc. With 1 and 7 as cen- 
tres and radius 01 describe a semicircle on 
opposite sides of the diameter; with centres 
2 and 6 and radius 02 do the same thing, 

and so continue taking each point as a centre and the distance 

from it to the end of the diameter as a radius. 




GEOMETRICAL PROBLEMS. 



41 



Prob. 76. To divide a circle into concentric rings having 

equal areas. 

Divide the radius DC into as many 
equal parts as areas required (Prob. 11) in 
points 1, 2, 3, etc. Ou CD as a diameter 
draw a semicircle, and at the points 1, 2, 
3, etc. draw lines perpendicular to CD, 
meeting the semicircle in points A, B, etc. 
With centre C and radii CA, CB, etc. 

draw concentric circles, which will divide the given circle as 

required. 




Prob. 77. A chord AB and a point C being given, to Jlnd 
other points in the arc of the circle passing through A, B, and C 
without using the centre. 

Draw AB, AC, and BC. Suppose 
four more points are required. With 
any radius and centres A and B draw 
the arcs DE and FG. Make CAE 
equal to CBA, and GBC equal to 
CAD. Divide the arcs DE and FG 
into the same number of equal parts, 
one more than the number of points required. Number the 
points 1, 2, o, 4, from D toward E, and from G toward F. 
Draw lines from A and B through these points, and those pass- 
ing through like-numbered points meet in points on the arc 
ACB. To construct a point on the curve below AB lay off 
GO equal to Gl on the arc FG, and D9 equal to Dl on the arc 
ED. Lines through and B, and A and 9 meet in K, a point 
on the curve. 




42 



GEOMETRICAL PROBLEMS. 




^. 



Prob. 78. On a chord AB to construct the supplementary 
arc to ACB, without using the centre^ C being a point on the arc. 

Join AC, AB, and BC. 
Make BAD and ABF each 
equal to ACB. With A and 
B as centres, and any radius, 
draw arcs HD and KF. Di- 
vide the arcs DH and KF 
into the same number of equal 
parts (say five) by the points 
1, 2, 3, etc. Number the 
points as shown. Draw lines from A through 1, 2, 3, etc. to 
meet lines through the same numbers drawn from B. The lines 
through like numbers meet in points on the required arc. 

Prob. 79. To construct any number of tangential arcs of 
circles^ having a given diameter. 

Suppose three (four) arcs are re- 
quired. Upon the given diameter 
as a base draw an equilateral triangle 
ABC (square ABCD). With each 
vertex as a centre, and a radius of 
half a side, draw arcs of circles tangent to each other, as shown. 

Prob. 80. At a 'point C on a line AB ^o draiv two arcs of 
circles tangent to AB, and to two parallels AD and ^^, form- 
ing an arch. 

Make AD equal to AC, and BE equal to BC. 
At C make CG perpendicular to AB, and at D and 
E draw the perpendiculars DF and GE, meeting 
CG in F and G, the required centres. 





GEOMETRICAL PROBLEMS. 



43 



Prob. 81. To draw any number of equidistant parallel 
straight lines hetiueen two given parallels, AB and CD. 

Draw any line Ao at any angle to AB, 
and lay off on it any convenient dis- 
tance Al as a unit as many times as 
lines required, plus one. Thus, if four 
parallels are required, lay off Al Jive 
times. With A as a centre and Ao as 
radius draw an arc cutting CD in point e. Join Ae. With 
centre A and radii Al, A2, A3, etc., cut Ae in a, h, c, etc. 
Through a, h, c, etc. draw parallels to AB and CD. 



A R 


1 ^A 


''' '\b 


2|-' \c 


.»i ^''"\d 


1 --' ^-e/ 


C 4- ^y - D 

si-'"'' 



Prob. 82. To draw through a point C a line to meet the 
inaccessible intersection of two lines, AB and DE. 

From C draw any lines CA 
~^ ^ and CD to the given lines. 

Join AD, AC, and CD. Make 
BE parallel to AD, BE par- 
allel to AC, and EF parallel 
to CD. The intersection of EF and BF gives a point F in 
the required line. Draw through C and F. 




Prob. 83. To draw a perpendicular to a line AB, which 

shall pass through the inaccessible intersection of two lines, AE 

and CD. 

Produce AB to cut the lines in A and C. 

From A draw a perpendicular to CD, and 
from C a perpendicular to AE ; these per- 
pendiculars meet in O. Draw BO perpen- 
dicular to AB. OB passes through the 
intersection of CD and AE. 




44 



GEOMETRICAL PROBLEMS. 



Prob. 84. To draw an involute of a square ABCD. 

Produce the sides as shown. With 
centre C and radius CD draw the arc DE. 
With centre B and radius BE draw the arc 
EF. Witli centre A and radius AF draw 
the arc FG, etc. 

Note. Suppose a line to be wrapped 
around and in the direction of the perim- 
eter of any plane figure. Let the line be unwound^ keeping 
it always straight in the process of unwinding. Any point in 
the line describes an involute. The involute of polygons is 
composed of arcs of circles, as in Prob. 84. 




Prob. 85. To draw the involute of a circle. 

Divide the circumference into any num- 
^ ber of equal parts, as at A, B, G, D, etc., 
and draw radii to these points. At A, B, 
C, D, etc. draw tangents. Let the curve 
start at A. On the tangent at B lay off a 
distance from B to 1 equal to one of the 
parts into which the circumference is di- 
vided. On the tangent at C lay off a distance equal to two 
parts to 2. On the tangent at D three parts to 3 ; from E four 
parts to 4, etc. The curve through these points, 1, 2, 3, 4, etc., 
is the involute of the circle. 




GEOMETRICAL PROBLEMS. 



45 




Prob. 86. To draw a spiral composed of semicircles, the 
radii being in arithmetical progression. 

Draw an indefinite line, BAG. On the line 
take any two points, A and B, as centres. 
With A as a centre and radius AB draw a 
semicircleo With B as a centre and radius BC 
draw a semicircle, and so on, using A and B 
as centres, and taking the radii to the end of 
the diameter of the last-drawn semicircle. 

Prob. 87. To draw a spiral composed of semicircles, whose 
radii shall he in geometrical progression. 

Let the ratio be 2. Let AB be the radius of 
the first circle, A its centre. Draw the semicir- 
cle BLC. With B as a centre and radius BC 
draw the semicircle CMD. With C as a cen- 
M tre and radius CD draw the semicircle DNE. 

D is the next centre, the diameter DE of the last-drawn circle 
becoming the radius for the next circle. So proceed. 




Prob. 



To draw a spiral of one turn in a given circle. 



Divide the circle into any num- 
ber of equal parts, say twelve, by 
the lines OA, OB, OC, etc , and a 
radius OA into the same number 
of equal parts by the points 1, 2, 
3, 4, etc. With O as a centre 
and radius 01 draw an arc cutting 
OB in M ; with centre O and radius 
02 an arc cutting OC in N ; with 
centre and radius 03 an arc cut- 
ting OD in point P, etc. Through M, N, P, etc. draw the curve. 
Note. This is " the spiral of Archimedes." 




46 



GEOMETRICAL PROBLEMS. 




Prob. 89. To draw in a given circle a spiral of any num^ 

her of turns, say two. 

Draw radii dividing the circle 
into any number of equal parts. 
Divide any radius, as OA, into as 
many equal parts as turns in the 
spiral, and divide each part into 
as many equal portions in points 
1, 2, 3, 4, etc. as the circle is di- 
vided into. With centre O and 
01, 02, 03, etc. as radii, draw- 
arcs to meet the radii OB, 00, 

OD, etc., respectively, in points of the required curve. 

Prob. 90. Given the axes AB and CD of an ellipse to draw 
the curve, and at any point on the curve to draw a tangent. 

Place the axes AB and CD 
at right angles to and bisecting 
each other at O. With centre 
C and radius OA cut AB in 
F and F', which are the foci. 
Between O and F', or F, take 
any point G, dividing AB into 
two parts. With centres F and 
F' and radius AG draw arcs on either side of AB. With the 
same centres and radius BG draw arcs intersecting those drawn 
with radius AG, at points L, M, N, and P, which are points on 
the curve. Take any other point, T on AB, and repeat the 
above operation ; and so on until as many points as are neces- 
sary are found. Through the points draw the curve. FM 
plus F'M equals AB. Let M be the point at which the tangent 
is required. Produce FM and draw the bisector of the angle 
SMF'. ME. is the tangent required. 



f-^" 


(? \V ^ 


\>:^ 





GEOMETRICAL PROBLEMS. 



47 



Note. For drawing the ellipse and similar curves through 
a series of points the so-called French Curves are to be used. 

Note. The major axis is sometimes called the transverse, 
and the minor the conjugate, axis. 

Prob. 91. To draw an ellipse by means of a trammel, the 
given. 

Let the semi-axes be OA and OB. Mark 
off on the straight edge of a slip of paper 
or card MP equal to OA, and NP equal to 
OB. Keep the trammel with the point N 
always on the major axis, and the point M 
on the minor axis, and P will be a point in 
the curve. Find as many points as necessary, and draw the curve. 




92. 



1 ' 1 "^nN 




/K-^. 




/ '^ ^-v-"- ^ 





;/.-' \ Aj 




( / 



Prob 92. To draw an ellipse, having given the axes. 

Let the semi-axes be OA and OB. With 
radii OA and OB and centre O draw circles. 
Draw any radii, OM, ON, etc. Make MP, 
NT, etc. perpendicular to OA, and HP, 
KT, etc. parallel to OA. P, T, etc. are 
points on the curve. 
/ 
Prob. 93. To draw an ellipse, having given the axes. 

Place the axes at rig-ht ano^les at their 
centres, and on them construct a rectangle, 
vr^-i^ Qiig iialf being shown in BDEC. Divide 
OA and DA into the same number of 
equal parts by points 1, 2, 3, etc. Draw 
lines through C and 1, 2, 3, etc. to meet 
lines from B drawn to 1, 2, 3, etc. on AD. 
P, Q, R, etc. are points on the curve. 






'v/ ^^'^"-.3,^''^ 






" c 


;' ■ --' - i 



48 



GEOMETRICAL PROBLEMS. 



Prob. 94. To draw a curve approximating to an ellipse. 

Draw the squares ABDC and BEFD, and 
their diagonals, intersecting in G and H. 
With centres G and H and radius GA draw 
the arcs AC and EF. With centres B and 
D and radius DA draw the arcs CF and AE. 




Prob. 95. To draw on a given line, AB as a major axis, a 
curve approximating an ellipse. 

Divide AB into three equal parts (Prob. 
11) by points C and D. With centres C and 
D and radius CA draw two circles intersect- 
ing in E and F. Through C and D draw 
ECG, EDH, FCK, and FDL, meeting the 
circles in points G, H, K, and L. With centres E and F and 
radius EG draw the arcs GH and KL, completing the curve. 




Prob. 96. Given the major and minor axes of an ellipse, to 
dratv the curve approximately. 

Let CA be the semi-major axis, and BC the 
semi-minor axis. Join A and B. Make CD 
equal to BC, and BE equal to AD. Bisect 
AE by a perpendicular, meeting BC pro- 
duced in F. With centre F and radius FB 
draw the arc BH, and with centre G and 
f}"' radius GH draw the arc HA. 

Note. One quarter of the whole curve is only shown, leav- 
ing to the student the construction of the full ellipse. 




GEOMETEICAL PBOBLEMS. 



49 




-i- 



Prob. 97. Having the axes given, to draw a curve of tan- 
gential arcs of circles approximating to the ellipse. 

AO is the semi-major axis, and OB 
the semi-miuor axis. Draw a rectangle 
with the axes as sides. AOBC is one- 
quarter of the rectangle. Draw AB. 
From C draw CMP perpendicular to 
AB, and meeting BO produced in P. 
Vj Make OE equal to OB. On AE as a 

\ diameter draw a semicircle AKE. Pro- 

duce OB to K. Make OL equal to BK. With centre P and 
radius PL draw the arc LN. Make AD equal to OK, and 
with centre M and radius MD draw the arc DN, meeting LN 
in N. Draw NMR and PNS. With centre M and radius MA 
draw AR ; with centre N and radius NR draw RS, and with 
centre P and radius PS draw an arc from S through B. Repeat 
in each of the quadrants. 



Prob. 98. To draio a parabola when the abscissa AB and 

the ordinate BC are given. 

Draw the rectangle ABCD, and divide 
AD and DC into the same number of 
equal parts. Through the points of divis- 
ion on AD draw parallels to AB, and 
from A draw lines to the points on DC. 
The first line above AB meets the line 
from A to the first point of division from 

D in a point P on the curve. The second parallel to A meets 

the second from A to DC and so on. P, Q, R, and C are points 

in the curve. Repeat the same below AB. 




50 



GEOMETRICAL PROBLEMS. 



Prob. 99. To draw a 'parabola when the directrix AC and 
the focus D are given, and to draw a tangent at any point L on 
the curve. 

This curve is such that its apex E 
is always half way between A and D, 
and the distance from D to any point 
upon the curve, as F, is always equal 
to the horizontal distance from F to 
the directrix. Thus DF equals FG, 
and DH equals HK, etc. Through 
D draw BDA perpendicular to AC. 
This is the axis of the curve. Draw parallels to AC through 
any points in AB, and with centre D and radii equal to the hori- 
zontal distances of these parallels from AC cut the correspond- 
ing verticals, which will give points on the curve. 

To draw the tangent at L. Draw the ordinate LN, meeting 
AB in N. Produce BA to the right. Make ET equal to EN. 
Draw LT, the tangent required. 




Prob. 100. To draw an hyperbola when the diameter AB, 
the abscissa BC, and the double-ordinate DE are given. 

Complete the rectangle 
BCDF, and divide CD and 
DF each into the same num- 
ber of equal parts. Draw 
BL, BM, and BN, inter- 
secting lines AK, AH, and 
AG respectively in points 
on the curve. Repeat below 
and in the other half of the 
curve as indicated. 





GEOMETKICAL PROBLEMS. 



51 



Prob. 101. 
circle. 



To draw an oval on the diameter of a given 



Let AB be the diameter. Draw the circle 
ACB. Make OC perpendicular to AB. Draw 
the lines BCD and ACE indefinitely. With 
centres A and B and AB as a radius draw the 
arcs BE and AD. With centre C and radius 
CD draw the arc DE. 

Note. The centres A and B may be taken anywhere on 
the line AOB produced. 




Peob. 102. 




K, which is the second centre. 
KMN. GN drawn from K as 



Upon a given line AB to draw an oval. 

Bisect AB at C, and draw 
the perpendicular CD, With 
B as a centre and radius AB 
describe an arc AD. Bisect 
the quadrant AE in F. 
Through F draw BFG. AG 
is the first part of the curve. 
Bisect CB in H, and draw 
HD. HD intersects BG in 
Bisect EL in M, and draw 
centre and radius GK is the 



second part of the curve. Bisect CH in 0, and draw DO. P 
the intersection of DO and KM is the third centre\ From P 
through E draw PET. NT drawn from P as a centre and 
radius PN is the third part of the curve. From E with a 
radius ET carry the curve to the line DC, and repeat the opera- 
tion for the other half of the curve. Draw a semicircle on the 
diameter AB for the other part of the oval. 



62 GEOMETRICAL PROBLEMS. 

CYCLOIDS, EPICYCLOIDS, ETC. 

The cycloid is the path described by any point in the circum- 
ference of a circle when the circle rolls along a straight line. 

An epicycloid is the path described by any point in the cir- 
cumference of a circle when the circle rolls along the outside of 
another circle. 

A hypocycloid is the path described by any point in the cir- 
cumference of a circle when the circle rolls along the inside of 
another circle. 

The rolling circle is called the generatrix, or generating cir- 
cle, and the line (straight or curved) on which it rolls is called 
the directrix. 

Every point in the tire of a wheel which rolls along the 
ground in a straight line describes a cycloid. 

Hence it is easily seen that in one revolution or turning 
around of the circle, or wheel, the circumference will roll out 
into a straight line, whose length will be equal to the circum- 
ference of the circle. To lay off the circumference on the 
straight line, either calculate its length or divide the circle into 
equal arcs, and lay off on the straight line as many divisions as 
there are arcs in the circle ; each division on the straight line 
being equal to the length of one division on the arc. 

The arc is more than the chord, so a distance greater than 
the chord must be taken, and this is obtained by judgment or 
by approximation. By dividing the arc into a number of smaller 
arcs, so that the chord practically coincides with the arc, the 
small chord being laid off as many times as there are small arcs, 
a length of straight line is obtained very nearly equal to the 
arc given. 



GEOMETBICAL PROBLEMS. 



53 




'W 



Prob. 103. To construct a cycloid. 

Let AB be the directrix, and AD 
the generating circle. Divide the 
rolling circle into any number of 
equal parts, say 12, and lay off 
these leno^ths of arcs alono; AB, 
giving points a, 5, c, etc. Through 
1, the centre of the generating cir- 
cle, draw a line parallel to AB. 
\ This is the line of centres. On 

this lay off 12, 23, 34, etc., equal to Aa, a5, etc., and with 
centres 2, 3, etc. draw the generating circle in all its posi- 
tions tangent to AB at points a, 5, c, etc. Draw through the 
points of division on the rolling circle parallels to AB, to meet 
the different positions of the rolling circle in points P, P, S, T, 
and U. These parallels are drawn in the figure from points 
jo, r, 5, ^, and u. Repeat the process for the other half of the 
curve. 

Another method is to take the chords of the arcs, B/>, Br, Bs, 
etc., and with centres a, 5, c, etc. cut the respective circles in 
points P, R, S, etc. The chord aP equals Bjt? ; JR equals Br; 
cS equals B5, etc. 

This method is better for the part of the curve APR, as the 
intersections are more clearly defined. 

Note. When the number of divisions of the rolling circle 
is large the curve may be drawn by arcs of circles by taking a 
as a centre, and radius aA, and drawing from A to P. Pro- 
duce Pa and R5 to meet, giving the centre for arc PR; R5 and 
Sc meet at the centre of arc RS, etc. 



54 



GEOMETRICAL PROBLEMS. 




Prob. 104. To construct an exterior epicycloid. 

Let R-A be the rolling circle on 
the outer circumference of the direct- 
ing circle. Divide R-A into any 
number of equal parts (say 12), and 
lay off these parts on Aah, etc., 
giving points a, b, c, d, etc. 

With the centre of the directing 
circle as a centre, draw an arc from 
R giving the line of centres R123 
etc. Draw from the centre of the 
directino; circle radial lines through 
a, h, c, d, etc., meeting the line of centres in points 1, 2, o, etc., 
the centres of the different positions of the rolling circle. With 
centres 1, 2, 3, etc. and radius RA draw the several positions 
of the rolling circle. With the centre of the directing circle 
as a centre, draw arcs through the points of division of the cir- 
cle R-A to meet the several positions of the rolling circle in 
points C, D, E, F, etc., which are points on the curve. Draw 
through C, D, E, F, etc. 

The points of the curve may be obtained by drawing from a 
as a centre and radius equal to the chord of one division of R-A 
an arc to meet the second position of the rolling circle in C ; 
from h with radius equal to the chord of two divisions an arc 
to meet the third position in D ; from c with radius equal to 
the chord of three divisions to meet the fourth position in E, 
etc. As in the case of the cycloid this method is better for the 
part of the curve ACD. Or from a lay off on the rolling cir- 
cle tangent at a one part of R-A (in this case ^^2) '■> ^^ the one 
tangent at h two parts of R-A ; on the one tangent at c three 
parts, etc. 



GEOMETEICAL PROBLEMS. 



55 



Prob. 105. To construct a hypocycloid. 




See the curve on the interior of 
the directing circle in figure. A«, 
ah, he, etc. are equal parts of the 
circumference of the rolling circle. 
R123, etc. is the line of centres. 
The process and directions, together 
with the letters and figures used, are 
the same as for the epicycloid. 

Note. When the diameter of the 
rolling circle is equal to the radius 
of the directing circle the hypo- 



cycloid becomes a straight line. 



Prob. 106. To construct an interior epicycloid. 

N \ Let the circle A-C roll on B-CMK 

On the circumference of B-C lay off 
equal arcs CM, MN, NO, etc. Draw 
from M, N, 0, etc. radial lines 
through B, and make MD, NE, OF 
each equal to the radius CA. With 
centres D, E, F, etc. draw the cir- 
cles tangent at M, N, O, etc. Lay 
off Ml equal to CM, giving point 1 ; 
from N lay off tivo divisions each 
equal to MC, giving point 2 ; from 

O lay off three spaces, giving point 3, etc. 1, 2, 3, etc. are 

points of the curve. 




B6 



GEOMETRICAL PROBLEMS. 



Prob. 107. To draw a scroll for a stair-railing. 

The circle ABCD is the eye of the scroll. 
Draw the diameters AC and BD at right 
angles. Draw the chord AD, and bisect 
it in 6. Draw a line 65 parallel to AC. 
Bisect AE in F. Bisect EF in 3. Make 
E4 equal to E3. Draw 32 and 45 paral- 
lel to BD, and 21 parallel to AC. From 
6 draw an indefinite line parallel to BD 
and produce 65, 21, etc. From point 1 
with radius IB draw the arc BH. From 
2 and radius 2H draw HJ, and so on. The arc BR of the inner 
curve is drawn from P with radius PB ; the arc RS is drawn 
from 6 with radius 6R. 




Prob. 108. 
heing given. 



To construct a spiral, its greatest diameter AB 



Divide AB into eight equal parts 
by points 1, 2, 3, etc. On 45 as a 
diameter draw a circle CDEF. This 
circle is the eye of the spiral. In- 
scribe a square CDEF, as shown in 
the enlarged drawing of the eye. Draw 
the central diameters of the square, 
12,10 and 11,9. Divide these diame- 
ters into six equal parts, and number 
as shown. With 1 as a centre and ra- 
dius IH draw an arc HK to meet a horizontal produced from 
2 through 1. With 2 as a centre and radius 2Iv draw the arc KL 
to meet a vertical through 3 and 2. With centre 3 and radius 3L 
draw arc LM meeting a horizontal produced through 4 and 3. 




GEOMETEICAL PEOBLEMS. 



57 



With 4 as a centre and radius 4M draw arc MN to meet a line 
drawn through 5 and 4 at N, and so proceed. 

The curve may be commenced by taking 12 as a centre and 
radius 12 A and drawinir from without toward the centre. 



Prob. 109. To describe an Ionic volute. 

Let AB be the vertical 
measure of the volute. Di- 
vide AB into seven equal 
parts, and from C, the lower 
extremity of the fourth di- 
vision, draw CF perpendic- 
ular to AB, of indefinite 
length. From any point on 
CF as a centre, with a ra- 
dius equal to one-half of 
one of the divisions of AB, 
draw the circle HIJK, form- 
ing the eye of the volute. Draw the diameter HJ perpendicu- 
lar to CF. Draw the square HIJK, bisect its sides, and draw 
the square 12 L M 11. Draw the dividing lines of the square 
as shown in the smaller figure, and extend them. The divis- 
ions corresponding to 12 N are equal. The divisions OP and 
RL are each equal to one-half of 12 N. From 1 as a centre 
and radius IH draw the arc HS ; from 2 as a centre and radius 
2S the arc ST ; from centre 3 and radius 3T the arc TU ; and 
so proceed in the order of the numerals. 

In drawing the inner curves the dots on the diagonals in the 
small figure indicate the centres. The division of the square 
of which 12 N is one side, shows how these centres are found. 




58 



GEOMETEICAL PROBLEMS. 



Fillet 



Bead 





Cavetto 



Torus 




Scotia 




Cyma Recta 




Cyma Reversa 



The Roman moldings are given above, the method of con- 
struction being evident. All the arcs are arcs of circles, and 
the angles are 45°, except in the flatter form of the cyma recta, 
where the line of centres is at 30°. 



CHAPTER III. 

INKING. 

15. It is supposed that the student has now become familiar 
with the use of the instruments necessary for the construction 
of a drawing in pencil, and has acquired a certain degree of 
proficiency in handling them which is necessary for accurate 
work. The next step is to learn to ink a drawing after it has 
been pencilled. 

Starting with a good pen, in good condition, and a smooth, 
well-ground black ink, it only remains for the student to learn 
to make a clean, sharp, even line. This may seem at first like 
an easy thing to do ; nevertheless, the ability to make a good 
ink line every time comes to most students very slowly, and 
after a great deal of practice. Therefore, before beginning to 
work on the plates, which are to be finished carefully and 
handed in for inspection, it will save a considerable time and 
paper to make lines against the triangle without regard to their 
length, direction, or location, until the student is thoroughly 
familiar with his pen and can make a fair line. Several hours, 
if necessary, in this preliminary practice will be invaluable. 

16. India Ink. A special ink, called India ink, is always 
used in making drawings. It comes either in the stick form, 
and has to be ground as used, or in the liquid form. In the 



60 INKING. 

latter form it is held in solution by an acid which corrodes the 
pen and eats into the fibre of the paper, so that if it is desired 
to erase a line it is much more difficult than if made with the 
ground ink. This ink is also very liable to rub off like soot. 
The only advantage it has is the saving of time in preparation. 
This kind of ink cannot be recommended for anything except 
coarse, rough work. It should not be used for tinting. 

To prepare the stick ink for use place a small amount of 
water in the ink slab or saucer (the slab should be perfectly 
clean), then grasp the ink firmly and, with a rotary motion, 
grind until the liquid is black and a little sluggish in its motion. 
After it has reached the point when it is black enough the 
grinding should cease, as a continuation only makes the liquid 
thicker, thus causing it to flow less freely from the pen. The 
liquid will look black in the slab after a very little grinding, but 
the necessary consistency will not be reached for some time. 
In order to determine when the proper point is reached make 
a heavy line with the drawing pen on a piece of paper and wait 
for it to dry : do not go over this line a second time. If the 
ink has not been ground sufficiently, it will look pale after it is 
dry, in which case more grinding is necessary. If at any time 
the ink becomes too thick, it can be diluted by putting in more 
water and mixing thoroughly. The stick* should always be 
wiped dry after using, to prevent its crumbling. The ground 
ink should be kept covered as much as possible to prevent evap- 
oration, which would soon cause it to become too thick for use. 
It is not advisable to prepare a large quantity of ink at once, 
as the greater the amount the longer it takes to prepare it, and 
freshly ground ink is preferable. If carefully covered, however, 
it may be kept two or three days. In case the ink becomes 
dry in the saucer it should all be washed out, as it is almost 
impossible to redissolve it entirely so that there will not be 



INKING. 61 

little scales which get into the pen and cause the ink to flow 
irregularly. 

17. Drawing Pen. This instrument, commonly called a 
right-line pen, is one of the most important of the drawing 
instruments, and it is very essential that it be of good quality. 
The screw is used to adjust the distance between the nibs, in 
order to make the line of the desired weight. The ink may be 
placed between the nibs by means of a brush or strip of paper, 
but it is more convenient to dip the pen into the ink, being 
careful to wipe the outside of the nibs before using. 

While inking the pen should be held so that both nibs rest 
on the paper evenly, and it should be inclined a little to the 
right, or in the direction of the line, with its flatter side against 
the triangle or straight edge, the end of the middle finger rest- 
ing against the side of the screw-head or against the side of the 
pen just below the screw-head. A slight downward pressure is 
necessary (the greater the rougher the paper), but do not press 
against the ruler, as the lines would be uneven in thickness. 
The ruler is simply a guide for the pen. The lines should 
always be drawn from left to right (relative to the person and 
not to the drawing). If it is desirable to go over a line a sec- 
ond time for any reason, it should be drawn in the same direc- 
tion ; never go backward over it. 

In inking a curved line by means of the right-line pen and 
irregular curve, it is necessary to constantly change the direc- 
tion of the pen so that the nibs shall always be tangent to the 
curve. This requires considerable practice to do nicely. 

In case the ink does not flow freely from the pen, moisten the 
end of the finger and touch it to the end of the pen, and try it 
on a piece of waste paper. If this fails the pen should be 
wiped out clean and fresh ink put in. In making fine lines the 



62 INKING. 

nibs of the pen are near to each other, consequently the ink 
dries between them quite rapidly, hence it will be found advis- 
able to clean out the pen thoroughly quite frequently to insure 
perfect lines. This is one of the secrets of being able to make 
good fine lines ; they should also be made more rapidly than 
heavy lines ; the heavier the line the slower the pen should be 
moved. Do not keep the point of the pen too near the straight 
edge, as the ink is liable to flow against it, thus causing a blot. 
Especial attention should be given to the care of the pens ; 
they should always be carefully wiped after using, and should 
not be put away with any ink dried on them, nor allowed to 
get rusty on the inside of the nibs. Any old piece of cotton 
cloth will answer to wipe the pens and stick of ink on. 

18. How TO Sharpen the Pen. To make good lines the 
pen must be kept in first-class condition, — that is, not only clean, 
but sharp ; and every draftsman should be able to sharpen 
his own pen. 

The curve at the point of the nibs of the pen should always 
be a semi-ellipse, with its long diameter coinciding with the axis 
of the pen ; it should not be a semicircle, nor should it be 
pointed. The student is advised to look carefully at the points 
of his new pen, so as to get a correct idea of the proper curve 
before it becomes changed by wear. When this curve becomes 
changed by wear, or if, from any other cause, one nib is longer 
than the other, the nibs should be screwed together, then, hold- 
ing the pen in a plane perpendicular to the oil-stone, draw it 
back and forth over the stone, changing the slope of the pen 
from downward and to the right to downward and to the left, 
or vice versa, for each forward or backward movement of the 
pen, so as to grind the points to the proper curve, making them 
also of exactly the same length. 



INKING. 6S 

This process, of course, makes the points even duller than 
before, but it is a necessary step. Next separate the points a 
little by means of the screw, and then place either blade upon 
the stone, keeping the pen at an angle of about 15° with the 
face of the stone, move it backward and forward, at the same 
time giving it an oscillating motion, until the points are sharp. 
This is quite a delicate operation, and great care should be exer- 
cised at first. The pressure upon the stone should not be very 
great, and it is well to examine the point very often so as to be 
sure and stop when each nib has been brought to a perfect edge, 
otherwise one nib is liable to be longer than the other and 
the pen will not work well, even if each nib is sharp of itself. 

Although the points want to be brought to a perfect edge, 
tEey should not be sharp like a knife, as in that case they would 
cut the paper. It will probably be necessary to try the pen 
with ink to be sure that it is in good condition. Sometimes a 
slight burr is formed on the inside of the blades ; this is re- 
moved by separating the points still farther, so as to insert the 
knife-edge of the oil-stone between them and draw it carefully 
through. One motion should be sufficient to remove the burr. 

The. pen should never be sharpened by grinding the inside of 
the blades other than as just indicated. 

19. Inking a Drawing. In inking a drawing it is prefer- 
able to ink all the circles and arcs first, as it is easier to make 
the straight lines meet the arcs than the reverse. Of a number 
of concentric circles the smallest should be inked first. Here, 
as in the case of the pencil compasses, the pen point should be 
kept nearly vertical, the top of the compass being inclined a 
little toward the direction of revolution, and there should be a 
slight downward pressure on the pen point, but none on the 
needle point. 



64 INKIKG. 

Where a large number of lines meet at a point care should 
be taken to avoid a blot at their intersection. These lines should 
be drawn from rather than toward the point, and each line 
should be thoroughly dry before another is drawn. 

In case of two lines meeting at a point neither line should 
stop before reaching the point, nor go beyond it. Either of 
these defects gives a very ragged appearance to the drawing. 

20. Stretching Paper. For ordinary small line drawings 
it is usually sufficient to fasten the paper to the board by means 
of thumb tacks, but for large drawings, or those which are to 
be tinted at all, it is necessary to stretch the paper by wetting 
it, and fasten it to the board with mucilage. To do thi^ lay the 
paper on the board, fold over about one-half an inch along each 
edge of the sheet ; next wet with a sponge the upper surface 
of the paper, except that portion folded over ; do not ruh the 
surface of the paper, simply press the wet sponge against it on 
all parts ; apply the mucilage to one of the edges and fasten 
that edge down, beginning at the middle and rubbing toward 
either end ; do the same with the opposite edge next, giving a 
slight pull to the paper as it is fastened down ; repeat this 
process for the two remaining edges. 

It is very important that the edges of the paper where the 
mucilage is to be applied should be kept dry, so that the muci- 
lage will be ready to act as soon as it can dry, and to facilitate 
this the less mucilage you can use and accomplish the result 
the better. If a large quantity of mucilage is used it will 
moisten the edges of the paper so much that it will be likely 
not to stick, as the body of the paper will dry as soon as the 
edges, and therefore pull them up. The drying of the mucilage 
can be hastened by rubbing the edge briskly with a piece of 
thick paper under the fingers until it becomes hot. The board 



INK^G. 65 

should never be placed near the fire or radiator to hasten the 
drying, as it would dry the paper before the mucilage set, caus- 
ing the edges to be pulled up. The board should be left to 
dry in a horizontal position, and all the superfluous water should 
be removed with a sponge, so as to avoid water marks in the 
paper, which always show in tinted drawings. In some cases, 
when the mucilage sets slowly, it may be necessary to moisten 
the centre of the paper sometime after stretching, to prevent 
its pulling up the edges by drying too rapidly. 

21. Correcting and Cleaning Drawings. Pencil lines 
are removed by means of a piece of rubber. When a mistake 
is made in inking, or it is desired to change a completed draw- 
ing, it becomes necessary to erase an ink line. This can be done 
by means of a rubber ink eraser, the same as in the case of 
pencil lines, except that much more rubbing is necessary. Ink 
lines can also be removed, and more quickly, by means of a 
knife ; in this case care should be taken not to use the point of 
the knife, as V-shaped holes are made which will always show. 
The flat portion of the knife should be used. After erasing an 
ink line, the surface which has been made rough by scratching 
should be rubbed down with some hard, perfectly clean, rounded 
instrument before inking other lines over it. 

A drawing can be cleaned by means of India rubber, or stale 
bread crumbled on the drawing and rubbed over it. Although 
dirt can be removed from a drawing, it should be the aim of 
the draftsman to keep it as clean as possible. Therefore, the 
drawing should be kept covered when not being worked upon, 
and, if the drawing is a large one, all except that portion which 
is in use should be kept covered. 



66 TINTING. 

TINTING. 

22. Tinting may be done in colors or India ink, as.desired. 
The method of putting on the tint is the same in either case ; 
consequently, we will take up only the India-ink tint here. 

If a drawing is to be tinted, the paper must be stretched as 
explained in Article 20. Especial care should be taken to 
keep the paper perfectly clean. That portion of the drawing 
which is to be tinted must not be touched with the India rubber, 
as the surface is thereby made rough and will not take a 
uniform tint. Hence, in laying out the work pencil lines must 
not be made on the surface to be tinted. 

23. Preparation of the Tint. Clean the ink slab, water 
glass, and the brushes thoroughly, also be sure that there are 
no scales on the stick of ink which could possibly come off. 
Fill the slab about half full of water, and grind the ink as pre- 
viously explained until it is black, but not thick. Fill the water 
glass about half full of clean water, and with a brush transfer 
enough of the ink in the slab to the glass to make a light tint. 
It is hard to get an ink which is absolutely/ free from specks, 
therefore, it is well to let the ink, after it is prepared in the 
slab, stand a short time to allow these specks to settle to the 
bottom ; then, in transferring the ink to the glass, do not plunge 
the brush down to the bottom of the slab, thus taking up this 
sediment, but let the brush fill from the surface of the liquid. 

The mixture in the water glass is the one to be used for 
tinting, and it is better not to make it as dark as you wish it 
upon the drawing when finished, as it is much easier to put on 
a light tint evenly than a dark one. The required depth of 
shade can be obtained by successive washes. Let each wash dry 
thoroughly before putting on another. A smoother effect can 
usually be obtained, especially on a large surface, by going over 



TINTING. 67 

the surface to be tinted with cleau water first, and then letting 
it dry. 

24. Laying on the Tint. Having laid out the surfaces 
to be tinted, incline the board so as to slope like an ordinary 
desk ; then dip your brush into the tint you have mixed, and 
take up as much of the liquid as it will carry, begin in the upper 
left-hand corner of the surface and draw it along the upper 
boundary, holding the brush nearly vertical, and leaving quite a 
puddle as you proceed. Lead this puddle gradually downward 
by going across the surface from left to right, about a quarter 
of an inch at a time, dipping the brush frequently in the tint, 
so as to keep the puddle about the same size all the time ; it 
should not be large enough to run down at any point. This 
puddle should not be left standing at any place any longer than 
is absolutely necessary, as it is very apt to leave a streak ; 
therefore, having commenced on a surface, finish it as quickly 
as possible, — do not let anything interrupt you. 

When you get to the bottom dry your brush on a piece of 
blotting paper, and with the dry brush take up the superfluous 
tint, as you would with a sponge, until it looks even. 

In laying on the tint do not bear on with the brush, as the 
brush marks would be liable to show, but use the point only, 
just touching it to the paper so as to wet it, and the tint will 
follow along of itself (the board being properly inclined). 

In following the boundaries of the surface to be tinted let 
the brush be pointed towards the boundary from the inside of 
the surface. 

This gives what is called a flat tint, and is used to represent 
surfaces which are parallel to the plane of projection. 

25. For representing surfaces which are oblique to the plane 
of projection a graduated tint is necessary. There are two 
methods of doing this, — the French and the American. 



68 TINTING. 

The French method consists in dividing the surface into small 
divisions (these divisions should be indicated only, not drawn 
across the surface), laying a flat tint on the first space, and 
when this is dry laying another flat tint on the first two spaces. 
Proceed in this way until the whole surface is covered, com- 
mencing at the first space each time. By this method the 
shading shows streaks of tint of different depths, but are almost 
unnoticeable if the divisions are taken quite small. 

The American method is most used, and is called shading by 
softened tints. 

There are two ways of doing this : — 

1st. By mixing a small amount of dark wash at first, and 
starting as if you were to put on a flat tint, and then, by re- 
peated additions of clean water, going over a little more surface 
at each addition, gradually make the dark tint lighter until you 
are using almost pure water. 

2nd. Divide the surface into divisions, as in the French 
method, only not so many ; put on a flat medium tint on the 
first space, but, instead of taking up all the tint from the bot- 
tom edge of the surface, leave a slight amount, touch the brush 
to some clean water and apply it to the lower edge of the pud- 
dle, thus making a lighter tint, and bring down this new tint a 
short distance ; repeat this a few times until the tint has prac- 
tically no color. Be careful to remove the most of the tint from 
the brush each time before touching it to the clear water. This 
work must be done even quicker than the ordinary flat tint. Use 
as little tint or water in the brush as you can and not have it dry 
in streaks. Let this dry, and then repeat the process, commenc- 
ing at the top and going over two spaces with the flat tint and 
softening off the lower edge, and so on, commencing at the top 
each time. Usually the tint should be softened out in the length 
of one division. If this shading is done perfectly, there will be 
a gradual change in the tint from beginning to end. 



CHAPTER IV. 



PROJECTIONS. 



26. Orthographic Projection, or Descriptive Geometry, 
is the art of representing a definite body in space upon two 
planes, at right angles with each other, by lines falling perpen- 
dicularly to the planes from all the points of the intersection of 
every two contiguous sides of the body, and from all points of 
its contour. 

27. These planes are called coordinate planes, or the planes 
of projection, one of which is horizontal, and the other vertical. 
H and V, Fig. 1, represent two such planes and their line of 
intersection GL is called the ground line. 

28. We shall only take, in this book, just enough of the 
elementary principles of projections to enable the student to 
make working drawings of simple objects. 

29. Since solids are usually made up of planes, planes of 
lines, and lines of points, if we thoroughly understand the prin- 
ciples involved in the projections of points, we ought to be able 
to draw the projections of lines, of planes, and of solids. The 
only difference being that, with a large number of points, the 
student is liable to get them confused. To avoid this liability 
it is advisable, at first, to number the points of a solid and their 
corresponding projections as fast as found lightly in pencil, and 
erase them after the problem is finished. Do not try to draw 

69 



70 PKOJECTIONS. 

the object all at once; it is impossible; one point at a time is 
all that can be drawn by anybody, and in this way the most 
complicated objects become simple, even though it may take a 
long time to complete the drawing. 

30. The projection of any point in space on a plane is the 
point at ivhich a perpendicular drawn from the given point to the 
plane pierces the plane. 

This perpendicular is called the projecting line of the point. 
Thus, in Fig. 1, a^ is the projection of the point a on the plane 
H, and a'^ of the same point on the plane V. These are called 
respectively the horizontal and vertical projections of the point 
a; aa^'- is called the horizontal projecting line of the point a, 
and aa^ the vertical projecting line of the same point. 

The horizontal projecting line aa^ is perpendicular to H, by 
definition, the plane V is assumed perpendicular to H, hence 
aa^^ is parallel to the plane V, and aa^ is equal in length to 
a%. Also the vertical projecting line, for the same reason, is 
parallel to H, consequently aa^ is equal in length to a^h. 

From the definition it is readily seen that each point in a line 
perpendicular to a plane will have its projection on that plane 
in one and the same point ; hence one projection of a point 
does not definitely locate its position in space. 

31. From the preceding article the following principles may 
be noted : — 

First, the perpendicular distance from the horizontal projec- 
tion of a point to the ground line is equal to the perpendicular 
distance of the point in space from the vertical plane ; or, 
briefly, the horizontal projection of a point indicates the distance 
of the point in space in front of V, but it conveys no idea of its 
distance above H. 

Second, the perpendicular distance from the vertical projec- 
tion of a point to the ground line is equal to the perpendicular 



PEOJECTIONS. 71 

distance of the point in space from the horizontal plane ; or, 
briefly, the vertical projection of a 'point indicates the height of 
the point in space above H, but it conveys no idea of its dis- 
tance in front of Y. 

32. If from the points a ^ and a'*, Fig. 1, perpendiculars 
should be erected to each coordinate plane, they will intersect 
at the point a in space ; and as two straight lines can intersect 
at only one point, there is only one point in space which can 
have al^ and a'^ for its projections. Hence two projections of 
a point are always necessary to definitely locate its position in 
space. 

33. It is evident that it would be very awkward to make 
our drawings on planes at right angles to each other ; hence 
the vertical coordinate plane is supposed to be revolved back- 
ward about its line of intersection GL with the horizontal 
plane until it forms one and the same surface with the hori- 
zontal plane, which may be considered to be the plane of the 
paper. 

In this revolution all points in the vertical plane keep the 
same distance from the ground line, and their relative positions 
remain unchanged. Thus, a^ revolves to a", a^h being equal 
to a^h, and as a^h was perpendicular to GL before revolution 
it will be so after revolution, and will form one and the same 
straight line with a%. 

Therefore, the two projections of a point must always he on 
one and the same straight line., perpendicular to the ground line. 

Now, if we draw a line across our paper and call it GL, all 
that portion in front of this line will represent the horizontal 
plane, and that portion behind it will represent the vertical 
plane, and the point a located as in Fig. 1 is represented by its 
projections on the plane of the paper as shown in Fig. 2. 



72 PROJECTIONS. 

34. A point situated upon either of the coordinate planes has 
for its projection on that plane the point itself, and its other pro- 
jection is in the ground line. 

This is readily seen by referring to Fig. 1 ; c^ and c^ are the 
projections of a point on H, and d^ and d^ of a point in V. 
These points are represented on the plane of the paper as 
shown by the same letters in Fig. 2. 

NOTATION. 

35. We will designate a point in space by a small letter, and 
its projections by the same letter with an A or ?; written above ; 
thus a^* represents the horizontal and a^ the vertical projection 
of the point a. This point may be spoken of as the point a, 
or as the point whose projections are a^a'^. 

The horizontal coordinate plane will be designated by the 
capital letter H, the vertical by the capital letter Y. 

Construction lines are those which are made use of simply to 
obtain required results. They are not a necessary part of the 
drawing, and when left on a drawing are intended to show the 
individual steps taken. 

To this end the student is expected to ink in all construction 
lines illustrating the special subject in hand until especially 
directed not to do so. That is, while on the subject of projec- 
tions it is not desired to have the geometrical construction lines 
inked in, but only those which refer to projections. When on 
the subject of shadows only those which show how the shadow 
is found are required. 

These lines should be inked in with a light, short dash not 
more than -^-^ of an inch long, and as light as the student finds 
he can make easily. 

All lines representing the projections of single lines, or edges 
of planes or solids, if visible, are inked in with a full, continu- 



PROJECTIONS. 73 

ous line, a little heavier than the construction lines ; if invisible, 
they should be made with short dashes the same length as the 
construction lines, but the same thickness as the visible lines, 
so as to distinguish them from the construction lines. 

The true length of a line when found should be inked in 
with a long and short dash, about the same thickness as the 
ordinary full line. When the true length of a line is given, it 
is put in like an ordinary construction line. 

Indicate an isolated point by drawing a small cross through it. 

In working drawings — which are practical applications of 
projections — horizontal projections are usually called plans, 
and vertical projections are called elevations. Therefore, they 
will be used synonymously throughout this book. 

The student should distinguish between the terms vertical 
and 'perpendicular. Vertical is an absolute term, and applies 
to a line or surface at right angles to the plane of the hori- 
zon while perpendicular is a relative term, and applies to any 
line or surface which is at right angles to any other line or 
surface. 

If one point is farther from Y than another, the first is said 
to be in front of the second point. Hence, if I say that a line 
slopes downward, backward, and to the left, it signifies that the 
line occupies such a position that the lower end is nearer V 
than the upper, and also that it is on the left of the upper end. 
Fig. 6 shows the projections of such a line. 

PROJECTIONS OF STRAIGHT LINES. 

36. A straight line is determined by two points, therefore 
it is only necessary to draw the projections of each end of a 
straight line and join them, and we have the projections of the 
line. If the line is curved, it becomes necessary to draw the 
projections of several points and join them with a curve. 



74 PROJECTIONS. 

37. If we lay a fine wire, ah, Fig. 3, on a horizontal plane, 
and also parallel to a vertical plane, its horizontal projection 
will be the wire itself, that is, a%^ is its horizontal projection, 
equal in length to the wire itself, parallel to GL, and at a dis- 
tance from it equal to the distance of the wire from V. 

The vertical projection of the end a will be at a^ in the 
ground line, and of h at If also in the ground line, since each 
end is on H (Art. 34). Hence the vertical projection of the 
line will coincide with GL between a" and b^, and a'^b'" is its 
vertical projection, a^b^ is equal in length to a^^b^, hence is 
equal to the actual length of the wire. 

Now, suppose the wire to be revolved from left to right about 
a horizontal axis, through the end a, keeping the line parallel 
to V. If a pencil were attached to the end b at right angles 
to the line, so that its point touched V at b^, it would trace a 
circular arc b'^c'^d", etc. (of which aP is the centre and a^5", or 
the true length of the line, is the radius) on the vertical plane 
as the wire is revolved ; the end a would, of course, not move. 
After the wire has been revolved through an angle of 30° its 
vertical projection will be at a^c", and it must be equal in length 
to the real length of the wire, a^ will be the horizontal pro- 
jection of the fixed end of the wire. Since the wire is parallel 
to V, every point in it is at the same distance from V, hence 
their horizontal projections must all be the same distance from 
GL, that is, in a line parallel to GL ; the horizontal projec- 
tion of the end c must also be in a line through c^ perpendicu- 
lar to GL, hence at c^ where this parallel and perpendicular 
intersect ; a^c^ then, is the horizontal projection of the wire 
after it has been revolved through an angle of 30°. 

For the same reason a^d^ and a^d^ are the two projections 
of the wire after being revolved through an angle of 45°. 

Similarly, a^e" and a^e^ are its projections after revolving 
through an angle of 60^. 



PROJECTIONS. 75 

When the line has been revolved through 90° it becomes per- 
pendicular to H, and its vertical projection is a^"^, perpendicu- 
lar to GL, and its horizontal projection is a point, a^, as might 
have been seen from the definition of the projection of a point. 

38. The same reasoning applies if we take the wire lying 
against V and parallel to H and revolve it about a vertical axis 
through the end a, as in Fig. 4. The projections are given for 
the dine lying against Y, and making angles of 30°, 45°, 60°, 
and 90° with V, being parallel to H in each position. 

39. The following principles may be noted from the pre- 
ceding articles : — 

1st. A line situated in either plane is its own projection on 
that plane, and its other projection is in the ground line. 

^nd. If a right line is perpendicular to either plane of pro- 
jection, its projection on that plane will he a point, and its pro- 
jection on the other plane will he perpendicular to the ground 
line and equal in length to the given line. 

3rd. When a line is parallel to either coordinate plane its 
projection on that plane will he parallel to the line itself and 
equal to the actual length of the line in space, and its projection 
on the other plane will he parallel to the ground line. 

4:th. If a line is parallel to hoth planes, or to the ground line, 
both projections ivill he parallel to the ground line, and equal to 
the actual length of the line. 

6th. If a line is ohlique to either coordinate plane, its pro- 
jection on that plane will be shorter than the actual length of the 



^th. If a line is parallel to one coordinate plane and ohlique 
to the other, its projection on the plane to ivhich it is parallel is 
equal to the true length of the line in space, and the angle which 
this projection makes with the ground line is equal to the true 
size of the angle the line in space makes with the plane to which 
it is ohlique. 



76 PROJECTIONS. 

1th. The projection of a line on a plane can never he longer 
than the line itself. 

Sth. If a point he on a line its projections will he on the 
projections of the line. 

9th. If a line in different positions makes a constant angle 
with a plane its projections on that plane will all he of the same 
lengthy without regard to the position it may occupy relative to 
the other plane. 

40. If two lines intersect in space their projections must 
also intersect, and the straight line joining the points in which 
the projections intersect must be perpendicular to the ground 
line ; for the intersection of two lines must be a point common 
to both lines, whose projections must be on the horizontal and 
vertical projections of each of the lines, hence at their inter- 
sections respectively. 

41. If two lines are parallel in space their projections upon 
the vertical and horizontal planes will be parallel respectively. 
If one projection only of two lines are parallel, the lines in 
space are not parallel. 

42. Any two lines drawn at pleasure, except parallel to each 
other and perpendicular to the ground line, will represent the 
projections of some line in space. 

43. Prob. 1. To draw the projections of a line of a defi- 
nite length and occupying a fixed position in space. 

Let it be required to draw the projections of a line 1" long, 
which makes an angle of 30° with H, and whose horizontal 
projection makes an angle of 45° with GL, the lower end of 
the line being ^" above H and J" in front of V. 

It is first necessary to place the line in such a position that 
its true leno^th and the true size of the ans^le it makes with one 
of the coordinate planes are shown, and these are only shown 



PEOJECTIONS. 77 

when it is parallel to one of the coordinate planes. In this case 
it must be placed parallel to Y, as the angle the line makes 
with H is shown in projection in its true size only when the 
line is parallel to V (Art. 39-6th). a? and a^ Fig. 5, are the 
projections of one end of the line, dP being ^" above GL and «^ 
^' below it. Through dP draw ayh"^ at an angle of 30° with 
GL and 1" long ; through a^ draw a%]'' parallel to GL, h]" be- 
ing found by dropping a perpendicular from h"^ (Art. 37). The 
two projections of the line, when parallel to V, are thus found 
to be dPl^ and al'lK 

Now let us suppose the end a of the line to be fixed and the 
whole line to be revolved through an angle of 45° about a ver- 
tical axis through this point, the line keeping the same angle 
with H. The horizontal projection will not change in length 
(Art. 39-9th), but will move through an angle of 45°, and will 
be found at aJ^}^. It is evident that in this revolution, so long 
as the angle with H does not change, every point in the line will 
remain at the same height above H. The point a does not move, 
being in the axis. We have seen that h^ moves to V^ ; If must, 
therefore, be somewhere on a perpendicular through h^^ and, 
since the points do not change their heights, it must also be on 
a line through 5 ^' parallel to GL, hence at their intersection If. 
Join dP and If and we have a^lf and aP-h^ as the required pro- 
jections of the line. 

44. If this line were revolved through 15° more, the point 
^ would go to c^, and If to c^, and oPc"^ and a^c^ would be the 
projections of the line making an angle of 30° with H, and 
whose horizontal projection made an angle of 60° with GL. 

If it were revolved still 30° more, the two projections would 
be aPdP and oP'd^^ each being perpendicular to the ground line. 
When a line is in this position, i. e., has its two projections in 
a line perpendicular to GL, it is said to be in a 'profile plane, 



78 PROJECTIONS. 

a profile plane being understood to be one that is perpendicular 
to both Y and PI. 

When a line is oblique to only one of the coordinate planes 
it is said to make a simple angle ; when it is oblique to both of 
them it is said to make a compound angle. 

45. If the angle that the line made with Y had been given, 
it would have been necessary to have first placed the line par- 
allel to H, and then to have revolved it about an axis through 
one end perpendicular to Y, in which case the length of the 
vertical projection would not change, and the points would not 
change their distances from Y. 

In Fig. 6, a^'b'" and a%^ are the two projections of a line 1" 
long, making an angle of 45° with Y, and whose vertical pro- 
jection makes an angle of 60° with GL. The principles and 
explanation for this construction are the same as for Prob. 1, 
if the horizontal and vertical planes are supposed to be inter- 
changed. 

46. Note. In drawing the projections of any line making 
a compound angle with the planes of projection, it is always 
first necessary to place it in such a position as will show its true 
length and the true size of the angle it makes with one of the 
coordinate planes, i. e., parallel to one of the coordinate planes. 

47. Prob. 2. To find the true length of a line given hy its 
projections, and the angle it makes with either plane of projec- 
tion. 

Let a^lf and a%^, Fig. 7, be the projections of the given line. 
The true length is only shown when it is parallel to one of the 
coordinate planes, hence this line must be revolved about an 
axis through either end until it is parallel to one of the planes. 
If it is revolved about a vertical axis through a until it is par- 
allel to Y^ the point a does not move, h^ moves to 5 ^ If is found 



PROJECTIONS. 79 

at h^ (where a perpendicular through hj^ intersects a horizontal 
through h^), and a^h^^ is the true length. Also, the angle which 
ayh^ makes with GL is equal to the true size of the angle the 
line makes with 11. 

If it had been required to find the angle this line made with 
V, it would have been necessary to have revolved the line about 
a horizontal axis until it became parallel to H. Assuming the 
axis through the end h, Fig. 7, a" moves to a ^, a^ to aj^, and 
aJ^W- is the true length of the line (which of course should equal 
a^hj-'), and the angle it makes with GL is equal to the true size 
of the angle the line makes with V. 

48. Note. The angles which the vertical and horizontal 
projections of a line make with GL are greater than the angles 
which the line in space makes with H and Y respectively, except 
when the line is parallel to one of the planes. 

49. Before studying farther the student should be sure that 
he thoroughly understands the foregoing principles, as the whole 
subject of projections depends upon them, and it is utterly use- 
}ess to try to go on unprepared. 

PROJECTIONS OF SURFACES. 

50. Plane surfaces are bounded by lines, therefore the prin- 
ciples which govern the projections of lines are equally appli- 
cable to these surfaces. 

51. If we suppose a rectangular card ahcd, Fig. 8, placed 
with its surface parallel to V and perpendicular to H, each edge 
being parallel to V, they will be projected on V in lines equal 
and parallel respectively, hence the true size of the card itself 
is shown in vertical projection. Two of the edges ah and cd, 
being perpendicular to H, are projected on that plane in the 
points a^ and d^ respectively. The other two edges, ad and be, 
are parallel to H as well as V, hence they will be projected in 



80 PEOJECTIONS. 

their true length on H, and, since one is vertically over the 
other, they will both be horizontally projected in the same line 

Now, if the card is revolved about one of its vertical edges 
as an axis, like a door on its hinges, the vertical edge which 
coincides with the axis does not move ;' the other vertical edge 
moves in the arc of a circle. The horizontal projection of the 
card will still be a straight line of the same length as before. 
Let the card be revolved through 60° ; aJ^ does not move ; d^ 
moves in the arc of a circle, of which a^ is the centre and a^d^ 
the radius, to d^; oJ^d^ is the horizontal projection of the card 
in its new position ; the vertical projection of the edge cd in 
this position is found at C^d^^ vertically above (f/*, and a^U'c^d^' 
is the vertical projection of the card after being revolved 
through an angle of 60°. 

If the card should be revolved through 30° more, i. e., 90° 
in all, its surface will be at right angles with both coordinate 
planes, and its two projections will be found at a"h^ and aV*, in 
one and the same straight line perpendicular to GL. 

52. If the card be placed on H, with one of its edges par- 
allel to y, a^'V'c^d^, Fig. 9, will be its horizontal and a^5^ its 
vertical projection. If this card be revolved about one of its 
edges which are perpendicular to V as an axis, like a trap-door 
on its hinges, through an angle of 30°, a^h^ and a^hj'-cl^d^ will 
be its two projections. If it be revolved through 60° more, or 
90° in all, its projections will be a"e'" and a^d^, which are just 
the same as a^'h^ and aV* in Fig. 8, as they should be, since the 
cards are the same size in the two figures and they occupy the 
same relative position in each, i. e., they are in a profile plane. 

53. The following principles may be noted : — 

Is^. When a plane surface is perpendicular to another plane 
its projection on that plane will he a line. 



PROJECTIONS. 81 

2w<^. Whe7i a plane surface is parallel to either coordinate 
plane, its projection on that plane will be equal to the true size 
of the surface and its other projection will he a line parallel to 
GL. 

drd. When a plane surface is perpendicular to one plane and 
oblique to the other, the angle which its projection on the plane to 
which it is perpendicular makes with the ground line is equal to 
the angle the surface in space makes with the plane to which it is 
oblique. 

4:th. If a plane surface, in different positions, makes a con- 
stant angle with a plane, its projections on that plane will all be 
of the same size. 

54. Prob. 3. To draw the two projections of a plane sur- 
face, or card, of a certain size, and making a compound angle 
with the coordinate planes. 

Let the card be of the size shown in Fig. 9, and suppose it 
to make an angle of 30° with H, and its horizontal projection 
an angle of 45° with GL. 

Draw the horizontal projection a^JV^c?^ of the card equal to 
its true size ; a^b'" will be its vertical projection. Revolve a"b^ 
through an angle of 30° to a^b^^, and a"hj^ will be the vertical 
projection of the card when it makes an angle of 30° with H 
and is perpendicular to V ; a^bj^cj'-d^^ is its corresponding hori- 
zontal projection. 

The angle with H is still to be 30° after the card has been 
revolved to its desired position, hence its horizontal projection 
will be the same size. Therefore, make a%^cj^d^\ Fig. 10, 
equal in size to d^bj^cM^, Fig. 9, and making the desired angle 
with GL. In this revolution, as long as one edge rests on H 
and the angle remains constant with H, every point keeps the 
same height above H, therefore the vertical projections of a^ 



82 PJtOJECTIONS. 

and d^\ Fig. 10, must be found at a" and c?^ ; also of b^ and c/* 
at b^ and c\ whose heights above GL are equal to the height 
of ^;, Fig. 9, above GL. 

The other parallelograms in Fig. 10 represent the projections 
of the same card at different angles with H, the horizontal pro- 
jections making the same angles with GL. 

55. a^V^c^kV^ and aPdyb^c^^ Fig. 11, represent the projections 
of the same card when it is lying on H with one of its diago- 
nals parallel to V, and d"d^b^'c^ and aP-b]^c]^d]^ are its projec- 
tions after being revolved about an axis perpendicular to V 
through the corner a through an angle of 45°. Fig. 12 repre- 
sents the projections of this card when, besides making an angle 
of 45° with H, the horizontal projection of the diagonal makes 
an angle of 30° with GL. The steps to obtain this are exactly 
the same as in Figs. 9 and 10, hence the explanation will' not 
be repeated. 

56. Pkob. 4. To draw the 'projections of a regular 'pentag- 
onal card, the diameter of the circumscribed circle being given, 
in two positions. 1st, when it is perpendicular to V and making 
an angle of 60° with H, one of its edges being perpendicular to 
V ; '2nd, when, besides making an angle of 60° with H as in 1st 
position, it has been revolved through an angle of 45°. 

The pentagon must first be drawn in its true size and posi- 
tion ; a%]^c]^d]^e^. Fig. 13, equal to the actual size of the card, 
is its horizontal projection, c V'' being perpendicular to GL, 
and a^b'^cf is its vertical projection. Revolve a^b'^c'^ through 
an angle of 60° to a^b^c'" ; each point moves in the arc of a cir- 
cle with a as a centre, and a^'b^c" will be the vertical projection 
of the card when in the 1st position asked for ; a'^b^c^d^'-e^^ is 
its corresponding horizontal projection. 

For the 2nd position revolve the plan just found through 45° 



k.2 



c* 



^* 



i<s^ 




Plate. I 










a 




1. 1 






^ 


i 
>» 



PEOJECTIONS. 83 

to the position a^'^b^, etc., shown in Fig. 14; d"h'"c^d^e^ will be 
the corresponding vertical projection. 

57. Cards of any shape and size, and occupying any position, 
may be drawn in the same way, care being taken to locate one 
point at a time. 

The first step must always he to place the card parcdlel to one 
of the coordinate planes. 

58. If the angle the card made with V had been given, it 
would have been necessary to have first placed the card parallel 
to V and then to have revolved it, through the angle it made 
with V, about an axis perpendicular to H, in which case the 
length of the horizontal projection, which is a straight line, 
would not change, and the several points would not change their 
respective distances from H. 

In the second revolution, which changes the angle with the 
coordinate planes from simple to compound, the vertical projec- 
tion must be revolved and the corresponding plan found. The 
angle with V being the same the vertical projection does not 
change its size ; the distances of the points in front of V remain 
the same after revolution as before, hence are found at the 
same distances from GL respectively. 

59. Prob. 5. To draw the projections of a circular card 
making a compound angle with the coordinate planes. 

Let the diameter of the card be given, the angle it makes 
with V, and the angle through which the vertical projection is 
to be revolved. 

A circle may be considered as a polygon of an infiniLe num- 
ber of sides, hence we can take as many points as we please on 
the circumference of the circle, and each one moves according 
10 the principles just described. 



84 PKOJECTIONS. 

Place the card parallel to Y ; a circle, a'^h^cj^ etc., Fig. 15, 
equal to the actual size of the given circle, is its vertical pro- 
jection, and al^h^c^^ etc. is its horizontal projection. 

Revolve the card through the required angle about a vertical 
axis through a ; a^V^c^, etc, is its horizontal, and a^Jfc", etc. is 
its vertical projection. 

The card is next to be revolved through a certain angle, still 
keeping the same angle with-V. The size of the vertical pro- 
jection will, therefore, not change. Hence, revolve the vertical 
projection found in Fig. 15 through the required angle to the 
position dPlPc'^^ etc., Fig. 16. None of the points change their 
distance from V, consequently ci^hf^c^, etc. is the horizontal pro- 
jection of the card, found as in the last problem. 

60. Fig. 17 shows a somewhat shorter method of drawing 
the projections of a circular card making a simple angle with 
the coordinate planes ; in this case it makes an angle of 30° 
with V and is- perpendicular to H. c^h^c^, etc., making 30° 
with GL, is its horizontal projection. Suppose the card revolved 
about its horizontal diameter ae until it is parallel to H. It 
will then be shown in its true size at a%c^, etc. ; hh^, cc^, etc. 
will show the actual distances of the points 5, c, etc. from the 
horizontal diameter ; d"e" will represent the vertical projection 
of the horizontal diameter about which the card is revolved. 
Of course, If must be found somewhere in a line through 6'*, 
perpendicular to GL, therefore, lay off t¥ equal to hh^, and b" 
is a point of the required vertical projection ; c'^s is made equal 
to c c'', d"r to dd^, etc. Other points may be found in the same 
way. 

It is evident that np^, mc^^ etc. are respectively equal to hh^, 
cc^', etc., hence it is only necessary to revolve the semicircle 
and the distance hh^ is laid off on both sides of the diameter 
qVqv^ giving the two points If and n". 



PEOJECTIONS. Sh 

PROJECTIONS OF SOLIDS. 

61. A cube is a solid bounded by six equal faces, and when 
it is placed so that two of its faces are parallel to H, and two 
others parallel to V, its two projections are dP¥e^f'^ and <ji^l^c^d^, 
Fig. 18. The top and bottom, being parallel to H, are hori- 
zontally projected in one and the same square, aJ^b^c^d^, which 
is, of course, equal to the exact size of these faces, and their 
vertical projections are a^h'" and e"/"" respectively (Art. 53-2nd) ; 
the front and back faces being parallel to V are vertically pro- 
jected in one and the same square, dPlfe^f'^^ which is also equal 
to the exact size of these faces, and their horizontal projections 
are aJ^V^ and c^d^ respectively (Art. 53-2nd) ; the left and right 
hand faces are perpendicular to both V and H, therefore their 
vertical projections are- dPe^ and 5"/^, and their horizontal pro- 
jections are a^d^ and h^c^ respectively (Art. 53-lst). 

The plan shows two dimensions of the cube, the length and 
breadth, and the elevation two dimensions, the length and thick- 
ness ; therefore, the three dimensions of the solid being shown 
in their true size in the two projections the object is completely 
represented. In this case it does not matter which projection 
is drawn first, as they each show two dimensions in their true 
size. 

62. Shade Lines. In outline drawings it is customary to 
put in shade lines, i. e., lines heavier than the others ; they give 
relief to the drawing, and, when properly placed, are of assist- 
ance in reading it. 

Shade lines, or edges, are those edges which separate light 
from dark surfaces (dark hy location and not by shadow cast). 

The rays of light are generally assumed to come from over 
the left shoulder in the direction of the diagonal of a cube, the 
person supposed to be facing the cube, and the cube to be in 
the position shown in Fig. 18. That is, the ray of light enters 



86 PEOJECTIONS. 

the cube at the upper, front, left-hand corner, whose projections 
are a" and a^, and leaves it at the lower, back, right-hand corner, 
whose projections are f^ and c^ ; the diagonal joining these 
points will represent the actual direction of the conventional 
ray of light, and its projections a^f'" and a'V^ are the projec- 
tions of this ray. The different rays of light are all supposed 
to be parallel to each other. 

It is evident from the figure that both projections of the rays 
of light make angles of 45° with GL. The student should 
distinctly understand that, although the projections of the ray 
of light make 45° with GL, the actual angle it makes with V 
or H is quite different. To find this angle apply the princi- 
ples of Art. 47 to Fig. 18, and we get a = 35° 15' 52" as its 
actual size. 

63. In the cube. Fig. 18, the top, front, and left-hand faces 
are light, and the bottom, back, and right-hand faces are dark, 
and the shade lines are, therefore, e^f"", which separates the 
front from the bottom, h'"f'", which separates the front from the 
right-hand face, V^c^, which separates the top from the right- 
hand face, and c^d^\ which separates the top from the back face. 
The two other shade edges which separate the left-hand face 
from the back and bottom faces are not seen in either projec- 
tion, since the top and front edges of the left-hand face are in 
the same plane and nearer the eye. 

It is for this same reason that the shade lines mentioned 
above are seen only in one projection, i. e., e""/^ is a shade line ; 
it is seen in elevation, but in plan is hidden by the upper front 
edge of the cube dPlf-d*'V^. 

64. It will be noticed that the right-hand and lower edges 
are shaded in elevation, and the right-hand and upper in plan. 
From this many draftsmen have adopted the arbitrary rule to 
shade the right-hand and lower lines in elevation, and the right- 



PROJECTIONS. 91 

pyramid. Joining 6° with a'^\lf,c^^ etc. we have its vertical pro- 
jection. 

The shade lines of a pyramid are not found directly by means 
of the 45° triangle, as we have been able to do previous to this, 
on account of the faces not being perpendicular to either coor- 
dinate plane. If we try to use the triangle as in the case of 
the prism, we would have said that the three faces, f^^o^^a^, 
a^^o^^V\ and h^o^^c^ were light, and the three remaining faces 
dark, but this is not the case. For let us suppose that the alti- 
tude of this pyramid is so small that each of the faces of the 
pyramid makes an angle with H less than 35° 16' (the angle 
the ray of light makes with H). It is evident that all of the 
sloping faces will be light, and the bottom being dark the shade 
lines would go entirely around the base. Now, if we consider 
the altitude to increase, we shall soon reach the point when the 
face o'^d'^e^ will become dark, all of the rest remaining light, 
and the shade line would change from d^^e^ to e^o^ and o^^d^^. If 
the altitude be still further increased, we next get the case 
shown in the figure where the face/^'oV* becomes dark, and the 
shade lines would change from e^^f^^ and e^o^ to f^^o^. If the 
altitude should be still further increased, the face c^o^d^^ would 
presently become dark also. 

Of course the other three faces would never become dark 
while the pyramid rested on its base, even if the vertex were 
extended to infinity, in which case we should simply have a 
prism. In cases like this, or where any surface is oblique to 
both Y and H, it is necessary to find the shadow of the object;' 
thus determining which surfaces are light and which are dark. 

71. In Fig. 24 a'^J'/^e^ is the elevation and cH'''c^'d^' is the 
plan of a rectangular prism, with two of its faces parallel to each 
of the coordinate planes. The plan shows its length and width. 



92 PROJECTIOKS* 

and the elevation its length and thickness. If a side elevation 
is desired, it will show the width and thickness. To get this 
the object must be projected onto a plane at right angles to the 
two coordinate planes, i. e., the profile plane, and this plane 
revolved about its intersection with V, as an axis, to coincide 
with Y. FOR is such a plane, resting against the end of the 
prism, PO being its intersection with Y and OR its intersection 
with H. 

In this revolution none of the points change their heights 
above H, nor their distances from the axis PO, hence the rect- 
angle 5^'cy]^m^^ will represent this side elevation, it being of 
course the same height above GL that the front elevation is, 
and the distance that c^^m^ is from the axis PO will be equal to 
the distance the back of the prism is in front of Y. 

The shade lines in the end elevation are shaded the same way 
as in the front elevation ; the ray of light is supposed to come 
from over the person's left shoulder when he is facing the pro- 
file plane, i. e., the vertical projection of the ray of light is the 
same for all elevations. 

72. Prob. 11. To draw the plan and two elevations of a 
square prism with its axis parallel to and at a definite distance 
from hath Y and H, all of its faces being oblique to both Y and 
H. Fig. 25. 

The end elevation is the only view of the prism which shows 
one of its surfaces in its true size and position relative to the 
coordinate planes, hence this view must be drawn first. 

Locate the point o'" at a perpendicular distance above GL 
equal to the height of the axis above H ; through o^ draw two 
lines at right angles to each other, making angles with GL equal 
to those made by the long faces of the prism with H respect- 
ively ; lay off on each of these lines, on both sides of o^, a dis- 



PEOJECTIOKS. 93 

taDce equal to half the side of the square ; through these points 
draw lines parallel to the lines through o^', and the square a^h^c^^ 
d'" thus formed will be the end elevation of the prism in its cor- 
rect position. 

To locate the axis the correct distance from V draw FOR, 
which represents the profile plane, perpendicular to GL and at 
a horizontal distance to the right of o^', equal to the distance of 
the axis in front of V. 

In the last problem the end view was constructed from the 
plan and front elevation ; in this pjoblem we construct the plan 
and front elevation from the end view by simply reversing the 
steps. 

The horizontal lines, a"e", ^'"f'"-, C^m", and d'"n'", drawn through 
a^^, h^^, c^, and d^ respectively, and each equal in length to the 
length of the prism, will be the front elevation of the different 
elements of the prism ; joining these ends the front elevation of 
the prism is complete. 

From O along OR lay off Om^, 0/^, etc. equal to the hori- 
zontal distances of c^"", h^^, etc. from FO. Through these points 
m^if^i etc. draw the horizontal lines m^c^^f^V^^ etc., each equal 
in length to the length of the prism, and joining the ends the 
plan of the prism, a^c^mJ^e^, will be complete. 

The long faces of the prism being perpendicular to V in the 
end view, the shade lines for that view may be found directly 
by using the 45° triangle, as shown by the arrows. In revolv- 
ing the prism from the position shown in end view to that in 
the front view, the front and back ends chano^e from light to 
dark and from dark to light respectively, but the long faces are 
light or dark in the front view and plan according as they are 
light or dark in the end view (provided the projection of the 
right-hand end is represented, which will be seen to the left of 
the front view). 



94 PROJECTIONS. 

73. Prob. 12. To draw the two -projections of a regular 
pentagonal prism, with its axis parallel to H and oblique to V, 
a7id its lower left-hand long face making a definite angle with H. 
Fig. 26. 

Here, as in the last problem, it is necessary to draw the view 
of the end of the prism when its axis is perpendicular to Y, so 
as to show it in its true size and position with regard to H. 
a^'h'^c'^'d'^e^ is its end view, the edo;e a'^h'" makino; an anoxic with 
GL equal to the angle the lower left face makes with H. In 
the last problem the prism was revolved through an angle of 
90° to its actual position, but in this it is revolved through a 
smaller angle. The steps being otherwise just the same, the 
explanation will not be repeated. 

The shade lines in this case may also be found, as in the last 
problem, by using the 45° triangle on the end view. 

74. Let us now suppose a case where the edge c^d^, Fig. 
26, makes an angle of 40° with H and in the same direction. 
It is evident that when the axis of the prism is perpendicular 
to V the surface which is projected in cl^d^^ will be light. Now, 
if the prism be revolved through 45° so that its axis makes an 
angle of 45° with V, in the same direction as shown in Fig. 26, 
it is also evident that the surface, which makes an angle of 40® 
with H, will now be dark, and the shade lines would therefore 
change. If the prism be revolved through 45° more in the 
same direction, its axis would be parallel to V, and the surface 
in question would then be light. That is, the surface when 
perpendicular to V would be light, but as it was revolved par- 
allel to H, at some intermediate position before it had revolved 
45°, it would become dark, changing to light again at some inter- 
mediate position between 45° and 90° of revolution. 

The same thing would occur if the face in question made any 



ia^- - 1 



Fig.23 




PlatG2. 




PROJECTIONS. 95 

angle with H between 35° 16' and 45°. The foregoing reason- 
ing would apply equally well to the under face a^h^^ except that 
this one would be dark where the corresponding upper one 
would be light. 

Therefore^ if a surface of a prism, as in the last 'prohlem, 
makes an angle with H hetween 35° 16' and 45°, that surface 
becomes doubtful in all its positions when the axis of the prism is 
oblique to V, and the shadow of this surface woidd have to be 
cast to determine positively whether it is light or dark. 

If the surfaces make angles with H, not included between 
the above limits, the 45° triangle on the end view would deter- 
mine the light and dark surfaces for all the oblique positions 
of the prism, as well as when the axis is perpendicular to V or 
parallel to V and H. 

75. Fig. 27 represents the two elevations and plan of a hol- 
low cylinder whose axis is parallel to V and H. Here the end 
elevation would naturally be drawn first, as in the last two prob- 
lems, but it is not strictly necessary, as both of its projections, 
when parallel to V and H, are the same, and the distance apart 
of the contour elements is equal to the diameter of the base. 

The student should note carefully the shade lines in the figure, 
especially in the end view. 

76. Fig. 28 shows the plan and two elevations of a pile of 
blocks. The lower one is a rectangular prism, the second one, 
which rests on the first, is the frustum of a square pyramid, and 
the top one is a square pyramid. In this case it is necessary 
to draw the plan first and construct the two elevations from it, 
according to the principles already explained. 

It will be observed that the group is considered as solid in 
putting in shade lines, i. e., the edges which represent the perim- 
eter of the base of the pyramid, for example, are considered 
as separating the sloping faces of the pyramid from the top 



96 PROJECTIONS. 

surface of the frustum on which it rests, and not from its base, 
as in Fig. 23. Compare the shade lines of the pyramids in 
Figs. 23 and 28. 

Since it is customary to tint drawings in which shadows are 
cast, shade lines would not be put in on the same drawing. 
Therefore^ it is advisable to disregard the shadows altogether in 
putting in shade lines on line drawings, i. e.,if a surface which 
would he light from its position is made darh because some other 
surface or body casts a shadow on it, treat the surface as light in 
putting in shade lines. 

11. Prob. 13. To construct the projections of a right hex- 
agonal prism when its axis raaTzes a compound angle with the 
coordinate planes, the angle it makes with H, the angle the hori- 
zontal projection makes with GL, and the size of the prism being 
given. Figs. 29, 30, and 31. 

There are three distinct steps necessary in the construction of 
this problem. 

Since the axis is oblique to both planes the prism cannot be 
drawn in the required position directly, but most be placed in 
such a position as will show two dimensions. Here the first 
step is to draw the two projections of the prism when its axis 
is perpendicular to II and its faces make the required angles 
with V (Art. 66). Fig. 29 represents the projections of the 
prism in this position. 

Next draw the projections of the prism after it has been 
revolved in the proper direction, so that the axis makes the 
correct angle with H, but is still parallel to V. Fig. 30 repre- 
sents the projections of the prism in this position. Since the 
prism keeps a constant angle with V, its vertical projection does 
not change its size (Arts. 39-9th and 53-4th). Hence, make 
the vertical projection in Fig. 30 the same size as in Fig. 29, 



PROJECTIONS. 97 

with its position changed so that the elements make the correct 
angle with GL. In this revolution no point changes its distance 
from V. Therefore, to construct the horizontal projection of 
the prism in this second position, draw through each corner, as 
dP^ a^, etc.. Fig. 30, lines perpendicular to GL until they inter- 
sect horizontal lines drawn through the corresponding points, 
as oP' in Fig. 29. This completes the second step. 

In the final position the axis is to make the same angle with 
H as it does in the position just drawn. Hence the horizontal 
projection must be the same size, however much it may be 
revolved. In this revolution no point changes its height above 
H. Therefore, draw the plan of the prism. Fig. 31, the same 
size as that in Fig. 30, only changing the angle the elements 
make with GL the required amount. Then from each corner 
of this plan, as <2^a/*, etc., draw perpendiculars to GL until 
they intersect horizontal lines drawn through the corresponding 
points, o^,a^^', etc., in Fig. 30. Joining these corners the ver- 
tical projection is completed. 

If the angle the prism made with Y and the angle the ver- 
tical projection made with GL had been given, the principles 
would have been just the same, only you would have first drawn 
it with its axis perpendicular to V, then revolved it about a 
vertical axis until it made the required angle with Y ; in this 
case the plan does not change its size ; and, lastly, revolved the 
vertical projection last found through the proper angle and con- 
structed the corresponding plan. 

The shade lines in Figs. 30 and 31 can only be determined 
positively by casting the shadows of the doubtful surfaces. 
Frequently it is possible to tell which are the light and which 
the dark surfaces without casting the shadows by conceiving the 
solid in its position in space together with the ray of light, but 
for the average student it would be little better than a guess 
until he has had considerable practice in finding shadows. 



98 PEOJECTIOKS. 

Before taking up these cases where it is necessary to cast the 
shadow in order to determine the shade line, it will be neces- 
sary to take up so much of the subject of shadows as will ena- 
ble the student to find the shadow of an ordinary object on the 
two coordinate planes. 

78. Prob. 14. To construct the projections of a right hep- 
tag onal pyramid when its axis makes a compound angle with the 
coordinate planes^ the angle it makes with H, the angle the hori- 
zontal projection makes with GL, and the size of the pyramid 
heing given. 

A careful examination of Figs. 32, 33, and 34 will be suffi- 
cient to understand this problem, since the principles are exactly 
the same as in the last problem. 

79. Prob. 15. To draw the projections of a prism 1" square 
and If" long, resting ivith one of its long edges on H, this edge 
making an angle of 60° with V, backward and to the right, its 
front end heing 2f" in front of Y., the loiver left-hand long 
face making an angle of 30° with H. 

Also, draw projections of a regular hexagonal pyramid whose 
altitude is 3", diameter of circumscribed circle about base is 1^". 
Tlie lower end of right-hand element of pyramid rests on H 1^" 
to the left of the point located in prism, and 1-|" in front ofY-, 
this element also rests on top edge of prism at a point ^' from its 
front end. The axis of pyramid slopes downward, backward, 
and to the left. The two lower faces of p>yramid make equal 
angles with H. Fig. 62 

The projections of the prism are drawn as already described 
in Art. 73 ; the spaces aP'd^, d^b^, and b^c^^ are made respectively 
equal to ad, db^, and bc^. 

Heretofore, in drawing the projections of an object making 



Fig 27 



Plata. 3. 




PROJECTIONS. 99 

a compound angle with the coordinate planes, we have had given 
the size of the object, the angle it made with V or H, and the 
angle its other projection made with GL, and the object has been 
drawn in three distinct positions. If the third position only is 
wanted it is not essential that the first two be wholly drawn, 
nor that they be made in separate figures. After the studenvi 
becomes familiar with this work, so that numerous lines do nol 
confuse him, a considerable part of the ^ construction may ha 
omitted. In the figure all the necessary construction lines have, 
been left in. 

Here neither the angle the pyramid makes with H, nor the 
angle its horizontal projection makes with GL, are given, but 
the projections of two points/ and e of one of the elements art, 
given, which enables the projections of this element to be drawu 
(indefinite in length). Revolve this line around until it is paralle'j 
to V, and lay off on it in this position the true length of the ele- 
ment. This line will, of coarse, show the true size of the angle 
this element makes with H, and the horizontal projection of 
the indefinite line before revolution shows the angle it makes 
with GL. All the necessary data are now obtained, and a care- 
ful study of the figure should enable the student to understand 
the rest of the construction. 

80. It is evident that neither the height of an object above 
H nor its distance in front of V affects, in the least, the size and 
shape of its projections. Therefore, the GL is not at all essen- 
tial in drawing the projections of an object unless its distances 
from V and H are given, which is not customary. In working 
drawings the GL is never used. It is only used in elementary 
projections as an aid in understanding the principles. 



CHAPTER V. 



SHADOWS. 



81. The shadow of a body upon a surface is that portion of 
the surface from which light is excluded by the body. 

The source of light may be assumed at any point, but it is 
customary to assume it so that the rays of light are parallel to 
the diagonal of a cube, as already stated in Art. 62, in which 
case its two projections make angles of 45° with GL. Although 
rays of light diverge in all possible directions from the source, 
yet when this source is as far removed as the sun there is no 
appreciable error in calling them all parallel. 

82. The shadow of a point on any surface is where a ray of 
light through that point pierces the surface. 

Hence, to find the shadow of a point on a surface, draw a 
line through the point to represent the ray of light, and find 
where it pierces the surface. 

83. To find the shadow of a point on H. 

In Fig. 35 let h represent the point in space and R the ray of 
light passing through this point. If and h^ will be the two pro- 
jections of the point h ; }fr and hH of the ray of light. 

The shadow of the point 5 on H will be where R pierces H. 
This point being in H will have its vertical projection in GL, 
and its horizontal projection will be the point itself (Art. 34) ; 
being in the ray of light its two projections must also be on 

100 



SHADOWS. 101 

the projections of the ray of light (Art. 39-8 th). Therefore, 
produce, if necessary, the vertical projection of the ray of light 
till it meets GL at r, and r will be the vertical projection of the 
point where the ray of light pierces H ; at r draw a perpendic- 
ular to GL and Vl^ where this perpendicular intersects the hori- 
zontal projection of the ray of light is its horizontal projection, 
and is the shadow of the point h on the plane H. 
This is shown in actual projection in Fig. 37. 

84. To find the shadow of a 'point on V. 

In Fig. 36 let a represent the point and R the ray of light 
passing through it. dP and a^ will be the projections of the 
point a ; aPr and aH of the ray of light. 

The shadow of the point « on V will be where R pierces V. 
This point being on V will have its ^on^OTZ^aZ projection in GL, 
and it will also be in the horizontal projection of R, hence at 
their intersection t ; the vertical projection of this point, and 
the shadow on Y, will be a^ where a perpendicular drawn from 
t to GL intersects oyr. 

This is shown in actual projection in Fig. 38. 

85. We have heretofore supposed that the coordinate planes 
did not extend below or behind their line of intersection, but 
they can just as well be considered as extending indefinitely in 
both directions, as shown pictorially in Figs. 35 and 36. Then, 
after the vertical plane has been revolved to coincide with the 
horizontal plane, that portion of the paper above GL represents 
not only that portion of V which is above H, but that part of 
H which is behind V; also that portion of the paper below GL 
represents that part of H which is in front of V and that part 
of V which is below H. 

Referring again to Fig. 35, we have already seen that R 
pierces H at Vl ; now, if we suppose R to be produced below H 
indefinitely, it must pierce V at some point, since it is not par* 



102 SHADOWS, 

allel to V. This point is found in exactly the same way as 
already described in Art. 84. It does not make a particle of 
difference whether it pierces V above or below H. That is, 
every ray of light, unless parallel to Y, will pierce Y at some 
point either above or below H, and since these points are all 
in Y their horizontal projections must be in GL. Of course, 
the shadow of the point h falls on H, and does not actually fall 
on Y, but the point can be found where it would fall if H were 
transparent, and it is frequently convenient to do this in finding 
shadows of bodies in certain positions, as we shall soon see. 

Fig. 37 shows this point h^ in actual projection. It being 
on that part of Y which is below H, after revolution appears 
below GL. 

Referring again to Fig. 36, it is evident that R not only 
pierces Y at «s, but also pierces H at a^. a^ being in H has 
its vertical projection in GL. 

Fig. 38 shows this point in actual projection. It being on 
that part of H which is behind Y, after revolution appears 
above GL. 

86. The following rules are evident from the foregoing : — 

To find the shadow of a point on Yi.^ produce the vertical pro- 
jection of the ray of light to meet GL ; erect a perpendicular at 
this point of intersection, and the intersection of this perpen- 
dicular with the horizontal projection of the ray of light will he 
the shadow required. 

It should be carefully borne in mind that this last perpendic- 
ular may intersect the horizontal projection of the ray of light 
above or below GL, depending on the location of the point in 
space ; that is, if the point be nearer H than Y the intersection 
will be below GL, and the shadow will actually fall on H ; if 
the point be nearer Y than H the intersection will be above 
GL, and the shadow will be imaginary. 



SHADOWS. ■ 103 

To find the shadow of a point on V, produce the horizontal 
projection of the ray of light to meet GL ; erect a perpendicular 
at this point of intersection, and the intersection of this perpen- 
dicular with the vertical projection of the ray of light will he the 
shadow required. 

Here, also, the same caution as for the last rule is applicable. 
But in this case if the point is nearer V than H the intersec- 
tion is above GL, and the shadow actually falls on V; if the 
point is nearer H than V the intersection will be below GL, 
and the shadow will be imaginary. 

The following may also be noted : — 

If the horizontal projection of the ray of light meets the GL 
before the vertical projection, the shadoiv will actually fall on V; 
if the vertical projection meets the GL before the horizontal, the 
shadow actually fcdls on H. 

87. Fig. 39 shows how to find the shadow of a line parallel 
to both y and H. 

Fig. 40 shows the shadow of a line perpendicular to and 
resting on H. 

Fig. 41 shows the shadow of a line perpendicular to H but 
not resting on H. In this case a part of the shadow falls on 
each of the coordinate planes. 

From these figures the following facts may be noted : — 

1st. The shadow of a straight line on a plane surface is a 
straight line. 

2nd. The shadow of a line on a, plane to which it is paral- 
lel is a line parallel and equal to it in length. 

3rd. The shadow of a line on a plane to which it is perpen- 
dicular coincides with the projection of the ray of light on that 
'plane, and (in case of the conventional ray^ is longer than the 
line itself. 

Ath. The shadow of a line on a plane may be said to begin 



104 SHADOWS. 

where the line pierces that plane, either or both being produced 
if necessary. 

5th. Since two points determine a straight line it is sufficient 
to find the shadow of two points of it on a plane surface. In 
cose the direction of the shadow, or the point where the line meets 
f^he plane surface receiving the shadow, is known, it is sufficient 
to construct the shadow of one other point only. 

^th. When the shadow of a line falls upon two surfaces which 
intersect, the shadows on the tivo surfaces meet at a common point 
on their line of intersection. This is equally true whether the 
two surfaces intersect at right angles to each other or otherwise. 

88. Prob. 16. To find the shadow on V and K of a line 
oblique to V and H, whe?i one end is nearer H than V, and the 
other is nearer V than H. Fig. 42. 

Let a^6^ and a^'-U^ be the two projections of such a line. The 
end a being nearer H than V its shadow will fall on H (Art. 
86), and will be found, as already described, at a^. The end b, 
being nearer V than H, its shadow will fall on V, and will be 
found at bl. Since the shadow of one end falls on H and of 
the other on V, it is evident that the shadow of the line will 
fall partly on H and partly on V, and also that a line joining 
these two points could only be a line in space, and therefore 
not the shadow required. 

It is essential that the two points which determine the shadow 
of a line should be on one and the same plane ; therefore, as we 
have the shadow of one point of the line on each coordinate 
plane, it is necessary to construct the shadow of another point 
of the line on either of the coordinate planes. Any point could 
be taken, but the ends being definitely projected it is more con- 
venient to use them. We have already found the shadow of a on 
H to be a^g ; the shadow of 5 on H is found at b^^, in the same 



SHADOWS. 105 

way. (This shadow we know does not actually fall on H, but 
it serves our purpose, which is to get the direction of the shadow 
of the line, just as well as if it did) ; 0.]^ is, therefore, the 
shadow of the whole line on H, but only that part of it, a!lc, 
which falls below or in front of GL is actual shadow. We 
have also seen that the shadow of 5 on V is l^, the shadow of a 
on V is similarly found at al, and aj5j is the shadow of the 
whole line on V, but only that part of it, UgC, which falls above 
or behind GL is actual shadow. We, therefore, have as the 
actual shadow of the line oJ^c on H and hlc on V, the portions 
on the two planes intersecting in a common point, c, on GL as 
already noted in Art. 87-6th. Since this is the case it is evi- 
dent that it is not necessary to find the shadow of the whole 
line on both V and H ; having found the point c where either 
shadow crosses GL, join this point with the shadow of the end 
which falls on the other plane. 

89. Fig. 43 shows the shadow on V of a square card whose 
surface is parallel to Y. Since the edges of the card are parallel 
to V their shadows will be parallel and equal each to each, and 
consequently the shadow of the card will be equal and parallel 
to the card itself. This will be true whatever the size or shape 
of the card. 

Therefore, the shadow of any plane figure on a surface to 
which it is parallel is a figure equal and parallel to it. 

90. Fig. 44 shows the shadow on V of a square card whose 
surface is perpendicular to V and parallel to H. Here the two 
edges ah and cd are parallel to V, consequently their shadows 
will be equal and parallel lines. The other two edges, ad and 
5c, are perpendicular to V, hence their shadows coincide with 
the vertical projection of the ray of light, consequently make 
angles of 45° with GL. 



106 SHADOWS. 

91. Fig. 45 shows the shadow on both Y and H of a square 
card whose surface is perpendicular to both V and H, that is, 
is in a profile plane. The edge cd is parallel to V, and is nearer 
to V than H, hence its shadow is on V parallel and equal to 
cd ; the edge ad is nearer to Y than H, and is perpendicular to 
Y, hence its shadow is on Y, and makes an angle of 45° with 
GL ; the edge ah is parallel to Y and perpendicular to H, the 
upper end is nearer Y than H, and the lower end is nearer H 
than Y, hence its shadow is partly on Y and partly on H ; that 
portion which falls on Y will be parallel to a^Jf, and that por- 
tion which falls on H will make an angle of 45° with GL ; the 
other edge he is parallel to H and perpendicular to Y, the front 
end is nearer H than Y, and the back end is nearer Y than H, 
hence its shadow will fall partly on Y and partly on H ; that 
portion which is on H will be parallel to h^c^, and that portion 
which is on Y will make an angle of 45° with GL. The 
shadow of the whole card on Y is a^hlcldl, of which only that 
portion alnmcldl, which is above GL, is visible. The shadow 
of the whole card on H is alblc'^d^, of which only that portion 
^mn is visible. It is, of course, not essential to find that por- 
tion a!lnmc\d^s of the shadow on H which is above GL. 

92. Fig. 46 shows the shadow of a card lying in a profile 
plane, all of its edges being oblique to both Y and H. Each 
point being found the same as all the preceding ones, no further 
explanation is necessary, 

93. Fig. 47 shows the shadow of a circular card parallel to 
H. The shadow on H we know must be a circle equal in size 
to the card, therefore it is only necessary to find the shadow of 
its centre o. This is found at o\. With this point as a centre 
and a radius equal to that of the card describe the arc of a 
circle mm. Since a part of the circle is above GL it is evident 
that that part of the shadow actually falls on Y. To get this 



SHADOWS. 107 

shadow take any points on the circle, as a,b,c,d, etc., and find 
their shadows separately, joining these points by a curved line. 
The points m and n, where the circle described from o^ as a cen- 
tre crosses GL, will, of course, be two points on the curve. 
This curve will be an ellipse, and the shadow of a circle on a 
'plane to which it is perpendicular or oblique will he an ellipse. 

94. As a solid is composed of planes, planes of lines, and 
lines of points, it is evident that the shadow of the most com- 
plex body is obtained by finding the shadow of one point at a 
time by the methods already given until the shadows of all the 
points on the object which cast shadows have been found, so 
that the student who finds himself now able to cast the shadow 
of any single point on a given plane has practically mastered 
the subject, and if such a one has any difficulty in finding the 
shadow of any object the trouble is that he does not understand 
thoroughly the subject of projections. 

95. Since the shade lines of a body separate its light from 
its dark surfaces, the shadow of the shade lines will form the 
boundary of the shadow of the body. Therefore, in finding 
the shadow of a body the shade lines should first be marked, 
if it is in such a position that they can be found by means of 
the 45° triangle, and the shadows of these lines give the shadow 
of the whole object. If the object is in such a position that 
the shade lines cannot be found by means of the 45° triangle 
directly, the shadow of every point on the object, except those 
which it is knowri do not cast shadows, should be found sepa- 
rately, and then join those points which will enclose the largest 
area. The shade lines can then be found from the boundary of 
the shadow by finding what lines on the object cast these boundary 
lines. 

96. Prob. 17. To find the shadow of an hexagonal prism 
with its two ends parallel to H. Fig. 48. 



108 SHADOWS. 

The shade lines of the prism in this position are found 
directly by means of the 45° triangle to be ah, he, cd, de, ef, 
fm, mn, and an. It is only necessary to find the shadow of 
each of these lines and join them in order, and the shadow is 
completed. The first three and last three of these lines are 
parallel to H, hence their shadows on H will be equal and^par- 
allel respectively to the edges casting them. 

97. Fig. 49 shows the shadow of a square prism on V and 
H, resting with its base on H and its long faces oblique to V. 
The shade lines are found first here. 

98. Fig. 50 shows the shadow of a cylinder on Y and H, 
with its base resting on H. The shadow of any number of 
points on the shade line between a and d can be found. That 
portion of the curve between aj ^^^ ^l will be a semi-ellipse, 
and the lines aim and din must be tangent to this ellipse at the 
points al and d^. 

99. Prob. 18. To find the shadoiv of a right cone resting 
with its hase on H. Fig. 51. 

It is evident that, unless all the sloping part of the cone is 
light, there will be two elements of the cone which separate 
light from dark surfaces, and also that these two elements meet 
at the vertex and terminate at the other end in the base. But 
we do not know just where the light surface stops and the dark 
surface begins, as we do in the case of the cylinder, so we have 
to cast the shadow first. 

The shadow of the vertex o is o^ ; from o'^ draw two lines 
o'^a^ and Ogh^ tangent to the base of the cone, and these will be 
the shadows of the two shade elements. We now see that oa 
and oh are the dividing lines between the light and dark sur- 
faces, but since these lines do not coincide with the contour 
elements od and oc, it is evident that neither od nor oc is a shade 
line. 




-^b'i I 



.(f 



Fig- 




PIoZg^. 




SHADOWS. 109 

The bottom of the cone is dark and the sloping surface o^aJ^c^^h^ 
is light, hence the edge a^c^b^ is a shade line. 

Note the difference in the shade lines on the cylinder and the 
cone. 

Do not shade the contour elements of a cylinder or cone. 

100. We have seen that to find the shadow of a body three 
things must be given, the body casting the shadow, the surface 
receiving the shadow, and the ray of light, and that these must 
be given by their projections. The surface receiving the shadow 
has for convenience so far been taken as one of the coordinate 
planes, one projection of which is the plane itself and the other 
is the ground line, but it is more frequently necessary to find 
shadows of objects upon themselves and upon other objects in 
the immediate vicinity. We have also seen that in, finding 
shadows on the coordinate planes we have only concerned our- 
selves with that projection of the surface which is a line, that 
is, the GL. Hence, to find the shadow of an object on surfaces 
other than the coordinate planes, we have only to find the line 
which corresponds with GL and proceed according to the rule 
already given (Art. 86). 

To find this line, which we may call GL, observe the follow- 
ing rule : — 

Rule for GL. The GL to he used in finding shadows is 
always that projection of the surface receiving the shadow which 
is a line. This line may be straight, curved, or otherwise. 

Of course, this rule is applicable only when one projection of 
the surface is a line, i. e., when the surface is perpendicular to 
one of the coordinate planes. 

lOL Prob. 19. To find the shadow of a stick of timber on 
the top and front of an abutment on which it rests. Fig. 52. 



110 ' SHADOWS. 

The shade lines of the stick of timber are easily found to be 
as follows: cd^. dd, de, and ef. First find the shadow on the 
top surface of the abutment ; the vertical projection of this sur- 
face is the line a^h'", which, according to the rule, is the GL to 
be used. The shadow of dd is d^'-d], of de is d^e^ and of ef 
(since ef is parallel to the top of the abutment) is a line through 
ej parallel to e^f^. Next find the shadow on the front surface 
of the abutment ; the horizontal projection of this surface is 
the line a%^, which must therefore be the GL for this surface. 

The shadow of the shade edge ef will fall on both the top 
and front surfaces, which intersect in the line ah^ therefore the 
shadows on the two surfaces will meet in a common point m on 
this line (Art. 87-6th). m" is, then, one point of the shadow 
of the edge ef on the front of the abutment. The point n'", for 
the same reason, is one point of the shadow of the edge cd^ on 
this surface. Since these two edges are parallel they will cast 
parallel shadows, hence it is only necessary to find the shadow 
of one more point, and the shadow is determined. The shadow 
of the point c is cl ; join this with n", and through m" draw a 
line parallel to CgW^, and the shadow is completed. Of course, 
any other point might have been taken on the edge cd^, or any 
point on the edge ef 

The irregular line r^s^ simply indicates that the abutment 
extends backward farther than it was necessary to show. The 
ragged end of the stick of timber indicates the same thing. 

102. Prob. 20. To find the shadow of a stick of timber 
on another stick of timber into which it is framed. Fig. 53. 

It is evident that the sloping stick will cast a shadow on the 
top of the horizontal stick. The lower front and the upper back 
edges cd and ab are the lines which cast the shadows, and mn is 
the line to be used as GL. The shadow of the point a is at 



SHADOWS. Ill 

Gg, apparently on the elevation of the object, but in reality it is 
where the shadow would be if the lower stick of timber were 
sufficiently wide to receive it. It is frequently necessary to 
imagine surfaces indefinite in extent for convenience in con- 
struction. The shadow of the edge ah begins where it leaves 
the surface (Art. 87-4th), therefore, join «J and h^ and that 
portion of the line which falls on the surface of the stick will 
be all that is necessary ; through c^, the point where the lower 
front edge cd pierces the top, draw a line parallel to h^a^, and 
the shadow is completed. 

103. Prob. 21. To find the shadow of one oblique stick of 
timber on another, both being 'parallel to V and lying against each 
other. Fig. 54. 

The shadow will fall on the front and top of the back stick. 
The ground line for finding the shadow on the front surface is 
rs, and for the top surface is mn. Here, as in Prob. 18, the 
horizontal projections of the points c" and c?", where the shadow 
leaves the front surface, will be points of the shadow on the top. 
A careful examination of the figure will make further explana- 
tion unnecessary. 

104. Prob. 22. To find the shadow of a straight wire 
lying on top of a vertical cylindrical wall on the wall and the 
shadow of the wall on itself Fig. 6b. 

Here the GL is a curved line, and is to be used just the same 
as heretofore. There being no new principles the student should 
be able to understand this problem from an examination of the 
figure. 

105. Prob. 23. To find the shadow of the head of a bolt 
on its shank when the length of the bolt is parallel to both V 
and H. 



112 SHADOWS. 

In Fig. dQ the head is hexagonal and the shank is cylindrical, 
and in Fig. 57 the head is octagonal and the shank is hexagonal. 

In cases like these it is evident that neither the plan nor ele- 
vation of the surfaces receiving the shadow is a line, but the end 
view, or the projection on a profile plane, is a line, and accord- 
ing to the rule is, therefore, the GL to be used, mno is the GL 
in Fig. 56, and mnor in Fig. 57. The only other new point to 
be noticed is that the two elevations of the ray of light each 
make an angle of 45° with a horizontal line, and slope in the 
same direction. 

106. Prob. 24. To find the shadow of a chimney or tower 
of a house on the roof. Fig. 58. 

The end view of the roof ran is the GL for this problem. 
The shadow is found the same as in Figs. 56 and 57. 

107. Prob. 25. To find the shadow of a stick of timber on 
the top and sloping faces of an oblique pier. Fig. 59. 

Since neither plan, elevation, nor end view of these sloping 
surfaces is a line our rule is no longer directly applicable, and 
we may make use of an indirect method (the shadow can be 
found directly by descriptive geometry methods). 

First find the shadow on top of the pier. The points c^ and 
e^, where it leaves the top, will be two points of the shadow on 
the left, front, sloping face (Art. 87-6th), c^ and e^ will be their 
vertical projections ; in the same way d^ and n^ will be two 
points of the shadow on the right, front, sloping face, and d^ 
and n" will be their vertical projections. The shadow of the 
lower front edge (the upper back edge could just as well have 
been taken) on the ground is the indefinite line f^^o^. The 
ground and the sloping face of the pier intersect, hence the 
point /^, where the shadow would go under the pier, will be a 



SHADOWS. 113 

point of the shadow on the left, front face ; f^ will be its verti- 
cal projection ; for the same reason o^ is a point of the shadow 
on the right, front face, and d" is its vertical projection. Join 
ey^, and through c^ draw a line parallel to it ; also, n^o^ and a 
line through d^ parallel to it. This completes the shadow on 
the plan. The shadovr on the elevation is found by projection 
from the plan. 

If the pier had been so high that the shadow on the ground 
would have been difficult to obtain, an imaginary horizontal 
plane could have been taken at any convenient place, and the 
shadow found on that, noting where it comes out from the pier. 
The dash and two dots line in plan and elevation represents the 
two projections of the intersection of such an imaginary plane 
with the pier. mH^ is the shadow of the lower front edge of 
the stick on this imaginary plane, which gives the points t^ and 
m^ as points of the shadow. The shadow is completed as before. 

108. Prob. 26. To find the shadow of any oblique line on 
any ohlique plane. Fig. 60. 

First find the shadow on any horizontal auxiliary plane. 
^vqv ^w\ \^q lY^Q vertical projection of one such plane, and will 
be the GL for this plane. The shadow of any point as a on 
this plane will be a^ ; the shadow of any other point could be 
found in the same way, thus getting the shadow of the whole 
line, but the shadow of a line begins on a plane where the line 
pierces the plane, and the line pierces this auxiliary plane at 
the point h (If being where m'^e'" intersects the vertical projec- 
tion of the line, and h^ being perpendicularly below it on the 
horizontal projection of the line), therefore, joining a^ and ^ 
we have the shadow on the auxiliary plane. m"e'" is the vertical 
projection of the line of intersection of the auxiliary plane with 
the card mnop ; its horizontal projection is, therefore, rnJ^e^. The 



114 SHADOWS. 

lines h^a!l and tinP-e^ lie in the auxiliary plane, the line m^e^ also 
lies in the plane of the card mnop, therefore, the point r^ where 
these two lines intersect, is a point on the card, and must be one 
point of the horizontal projection of the shadow required ; r^ on 
mV is its vertical projection. Assume any other auxiliary 
plane, as d^o'", and another point, s'"s^ of the shadow, will be 
found in the same way. Draw an indefinite line through these 
points r and s in both projections, and the shadow is finished. 

Vertical auxiliary planes could have been taken instead of 
horizontal with the same result. 

109. Prob. 27. To cast the shadow of an abacus on a con- 
ical column. Fig. 61. 

Since neither projection of the surface receiving the shadow 
is a line, this problem must be done by the indirect method as 
used in the two preceding problems. 

Find the shadow on any horizontal plane, as a^c'". Its ver- 
tical projection is a"c^, and is the GL for this shadow. This 
plane cuts the column in a horizontal circle, of which a^c'" is 
the vertical and d^V^c^ is the horizontal projection. The shadow 
of the bottom edge of the abacus on this plane is the circle 
d^a^e^, drawn with o^, the shadow of o the centre of the abacus, 
as a centre and radius equal to that of the abacus. This circle 
cuts the circle a''6V* at the point a^\ which is one point of the 
horizontal projection of the shadow required, a^ on the verti- 
cal projection of this auxiliary horizontal plane, is one point of 
the vertical projection of the required shadow. Any number 
of other points can be found in the same way. 

110. Prob. 28. To cast the shadow of the prism given as 
in Fig, 62 07^ H and V, also of the pyramid on the prism and 
on H. 




Fig.60 




Plates. 




SHADOWS. 115 

The shadows of the prism on H and Y and of the pyramid 
on H require no explanation. 

If the shadow of a line falls partly on two plane surfaces, 
A and B (no figure), which do not intersect (at that portion 
under discussion), A being between the line and B, that part 
which falls on A does not fall on B, and the shadow of the line 
on B may be said to begin at the shadow of the point where 
the shadow of the line leaves A, on B. Now, if B is horizontal 
or vertical, and A is oblique to both V and H, the shadows of 
the line and the plane A on B can be readily found ; and then 
get the shadow of the line on A by the reverse of the above 
process, that is, note where the shadow of the line and plane A 
on B intersect ; find what point on A cast this intersecting 
point, and that will be one point of the shadow of the line on A. 

Now, referring to Fig. 62, we see that the shadow of the 
pyramid and prism on H intersect at the four points r^', fg, oc^, 
and Zg ; the points on the prism (this being between the pyra- 
mid and H) which cast these shadows are r^, t^, x^, and z'^ 
respectively (found by drawing a 45° line from the shadow to 
the edge casting it), and they are, therefore, points of the 
shadow of the pyramid on the prism. 

The shadow on the upper edge of the prism may be found 
by this same principle, that is, the shadow of this upper edge 
on H is h}^/^ ; this intersects the shadow of the pyramid on H 
at Sg and y^ ; s^ and y^ are the points which cast these shadows, 
hence they are points of the shadow required. Join /'s'*, sH'^, 
a^y, and y^z^, and the horizontal projection of the shadow is 
completed. Its vertical projection is found by projecting each 
one of these points on to the vertical projection of the prism. 
The points s^ and y* could also have been found by casting the 
shadow of the pyramid on an auxiliary horizontal plane through 
the top edge of the prism ; o js'* and o ^^ are the shadows of 



116 SHADOWS. 

the two shade elements of the pyramid on such a plane ; they 
intersect the upper edge of the prism at the points s^ and y^, 
which are the points required. A vertical auxiliary plane 
through this edge or the other edges of the prism might have 
been used with the same result. 



CHAPTER VL 

ISOMETRICAL DRAWING. 

111. In all the previous constructions two projections have 
been used to represent a body in space. In isometrical projec- 
tions only one view is used, the body being placed in such a 
position that its principal lines or edges (length, breadth, and 
thickness) are parallel to three rectangular axes, which are so 
placed that equal lengths on them are projected on the plane 
equal to each other. Thus we have the three dimensions of a 
body shown on one plane in such a way that each can be meas- 
ured, thereby combining the exactness of ordinary projections 
and the intelligibleness of pictorial figures. It is used chiefly 
to represent small objects in which the principal lines are at 
right angles to each other. In large objects the drawing would 
look distorted. 

112. If we take a cube situated as in Fig. 63, and tip it to 
the left, about its lower left corner e until the diagonal eg is 
horizontal, Fig. 64, and then turn it through an angle of 90°, 
still keeping eg horizontal, we obtain Fig. 65. The vertical pro- 
jection in this figure is what is called an isometrical projection. 

The edges of a cube are all of equal length, and it will be 
seen that they all make equal angles with the plane, hence they 
will appear equal in projection, consequently the visible faces 
must also appear equal. It will also be seen that the figure 



118 ISOMETEICAL DRAWING. 

can be inscribed in a circle, and that the outline of the isomet- 
rical projection is a regular hexagon, hence, that those lines 
which represent length and breadth make angles of 30° with a 
horizontal, and those which represent thickness are vertical. 

113. The edges of the cube being inclined to the plane on 
wdiich they are represented appear shorter than they actually 
are on the object, but since they are all equally foreshortened, 
and since a drawing may be made to any scale, it is customary 
to ignore this foreshortening, and make all the isometrical lines 
of the object equal to their true lengths. This will give what 
is called the isometrical drawing of the object, which will be 
somewhat larger than the isometrical projection. 

The vertical projection of the cube as shown in Fig. 65 
represents the isometrical projection of the cube shown in Fig. 
63, and Fig. 67 is the isometrical drawing of the same cube. 

114. Definitions, c'", Fig. 65, is the isometric centre. 
c^lf^ c^'d^\ and c^e^ are the isometric axes. Lines parallel to 
either of the isometric axes, or edges of the cube, are isometric 
lines, while any line not parallel to one of these axes is a non- 
isometric line. Planes parallel to the faces of the cube are 
isometric planes., and those which are not parallel to one of these 
faces are non-isometric planes. 

115. Direction of the Rats of Light. In isometrical 
drawing the rays of light are supposed to be parallel to a diagonal 
of the cube, as in ordinary projections, only here it is parallel 
to the plane of projection, and it is represented by a line making 
an angle of 30° with a horizontal line. Any line parallel to 
df, Fig. 67, may represent a ray of light. 

116. Shade Lines. These are the same as in ordinary 
projections, that is, they separate light from dark surfaces. In 



ISOMETKICAL DRAWING. 119 

all rectangular objects the top, left front, and left back surfaces 
are light, while the bottom, right front, and right back surfaces 
are dark. In Fig. 67 the edges ah, he, ce, and el are the visible 
shade lines. And all rectangular objects have their shade lines 
in relative positions to those of the cube. As in ordinary pro- 
jections, in putting shade lines on a group of objects touching 
each other, the group is shaded as if it were one solid ; also, in 
outline drawing the shadows are disregarded in putting in shade 
lines. 

117. Prob. 29. To make the isometrical drawing of a cuhe. 
Fig. 67. 

With the centre c and radius equal to the edge of the cube 
draw a circle, and in it inscribe a regular hexagon, and draw the 
alternate radii ch, cd, and ce, and the drawing is completed. 

Another method, which is applicable to any rectangular ob- 
ject, is to draw from any point as e lines ef, and el, each mak- 
ing an angle of 30° with a horizontal, and the vertical line ce; 
on these lay off ef, el, and ec equal to the true length of the edge 
of the cube ; from the points f and I draw indefinite vertical 
lines ; from c draw the lines ch and cd parallel to ef and el, 
intersecting the verticals through f and / in the points h and d ; 
from the points h and d draw lines parallel to cd and ch, meet- 
ing in a. This completes the drawing. 

118. Prob. 30. To make the isometrical drawing of a rect- 
angular hlock, ivith another rectangular hloch resting on its top 
face, and a recess in its right front face. Fig. 66. 

Construct the isometrical drawing of the large block by the 
second method of Art. 117. 

To draw the small block it is first necessary to locate one of 
its lower corners in the top face of the large one. This must 



120 ISOMETEICAL DRAWING. 

be done by means of two coordinates referred to two isometric 
lines as axes. The point e, in the top face of the large block, 
is -J" from the side and ^V" from the end, therefore make af 
equal to ^' and ag equal to ^", then from f and g draw the 
isometric lines ef and ge^ intersecting in e, the point required. 
The rest of the small block is drawn in the same way as the 
large one. 

To make the recess in the front side the point t is located 
in the same way as the point e, and tx equal to the depth of 
the recess is laid off as shown. The rest of the construction 
is evident. 

119. Fig. 68 shows the isometrical drawing of a rectangular 
box, without cover, 15" long, 6" wide, and 4" high, outside 
measurements, the boards being f" thick. The scale being 
1|^"— r. The ends are nailed on to the sides, and the bottom 
is nailed to the sides and ends. The visible joints are shown. 
The dotted lines show the inner edges of the box which are not 
visible. 

120. Fig. 69 shows the isometrical drawing of a four-armed 
cross. A careful examination of the figure will enable the 
student to understand its construction, there being only iso- 
metric lines involved. 

121. Prob. 31. To make the isometrical drawing of the 
pentagonal prism shown in Fig. 20. Fig. 70. 

The edges of the base, not being at right angles to each other, 
are non-isometric lines, hence the base should first be inscribed 
in a rectangle. Let one side of the rectangle contain the edge 
be, and the other sides respectively contain the corners a, d, 
and e. Make the isometrical drawing of the rectangle and 
locate each corner of the base, a, b, c, d, and e, by laying off on 



Fig.66 




PlatG 6. 




ISOMETKICAL DEAWIXG. 121 

the sides of this rectaDgle the distance of each point respect- 
ively from the nearest corner of the circumscribed rectangle. 
At these points draw vertical lines, and lay off on each of them 
the true height of the prism and join the tops, completing the 
drawing of the prism. It will be noticed that the non-isometric 
lines on the drawing are not equal to their true length on the 
object. 

122. Prob. 32. To make the isometrical draiving of an 
oUique timber framed into a horizontal one, as given in Fig. 53. 
Fig. 71. 

The horizontal timber is drawn as usual. To draw the oblique 
one, these edges being non-isometric, two points have to be 
located by means of coordinates. The point a is located by 
making ad equal to the distance the lower end of the oblique 
stick is from the end of the horizontal one. Any other point 
h is found by making cd equal to the horizontal distance of the 
point h from d, and he equal to its vertical distance from d. Join 
ah, which gives the isometrical drawing of one edge of the ob- 
lique timber. The other edges are, of course, parallel to this, 
being drawn through the points e and f which are located the 
same as point e in Prob. 30. 

123. Prob. 33. To make the isometrical draining of the 
skeleton frame of a hox made in the form of the frustum of a 
square pyramid. Fig. 72. 

Let a square prism be circumscribed about the frustum. The 
isometric of this prism is readily drawn, and is shown by the 
dotted lines. The bottom edge of the frustum coincides with 
the bottom of the prism.- The points «, 5, c, and d are in the 
upper face of the prism, and are found as the point e is in 
Prob. 30. 



122 ISOMETEICAL DRAWING. 

Join ah, he, cd, da, af, hg, and de, and the main frustum is com- 
pleted. The other lines which change the frustum from a solid 
to a skeleton B@ed no explanation. 

124. Prob. 34. To make an exact isometrical drawing of 
a circidar card, and also of any scroll or letter on its surface. 
Fig. 74. 

Let the circle and letter G be given as in Fig. 73. First, 
circumscribe a square about the circle. Make the isometric 
drawing of the square. The centres of the sides of the square, 
d, e,f and g, give four points of the isometric of the circle. 
Each point on the circumference of the circle, as a, h, c, etc., 
has two coordinates, by means of which the isometrical draw- 
ing of the points may be easily obtained. There will be four 
points on the circle whose coordinates will be the same as those 
for c, and eight points whose coordinates will be the same as 
those for a or h, or any point between d and e. The more 
points that are taken the more accurate will be the ellipse which 
forms the isometrical drawing of a circle. 

The letter is drawn in the same way by taking the two coor- 
dinates of several points on it. 

125. Prob. 35. To make an approximate construction of 
the isometrical drawing of a circle. Fig. 75. 

Make the isometrical drawing of the circumscribed square as 
before; d, e,f, and g will be four points. Draw the lines ag, 
af hd, and he, intersecting in the points c and o ; c will be the 
centre of the arc between d and g ; o of that between e and f; 
a of that between g and/; and h of that between d and e. 

126. Fig. 76 shows the approximate construction of the 
isometrical drawing of circles in each of the three visible faces 
of a cube. No explanation is necessary. 



ISOMETEICAL DK AWING. 123 

The student should study this and Fig. 75, so as to be able to 
make the isometrical drawing of a quarter of a circle in either 
isometric plane without making the whole circle. 

127. Fig. 78 shows the isometrical drawing of a bolt, an 
hexagonal nut, and a circular washer as shown in Fig. 77. 

128. Fig. 79 shows the isometrical drawing of the hollow 
cylinder given as in Fig. 27. 

129. Prob. 36. To divide the isometrical drawing of a 
circle into equal parts. Fig. 74. 

At the middle point f of one of the sides nl of the isomet- 
rical drawing of the circumscribed square draw/o perpendicu- 
lar to nl, and make/b equal to the radius of the circle ; draw 
ol and on; with o as a centre and radius ^b describe the arc 
rfs ; divide this arc into any number of parts ; draw through o 
and these points of division lines to meet nl ; join these meet- 
ing points with t, the centre of the ellipse, and where these lines 
intersect the ellipse will be the points of division of the isomet- 
rical drawing of oue quarter of the circle. 

The other quadrants can be divided in the same way. 

Another Method. On the long diameter of the ellipse 
draw the semicircle chp ; divide this into any number of equal 
parts ; through these points of division draw lines perpendicular 
to cp ; where these lines intersect the ellipse will be the points 
of division sought. 

130. In isometrical drawing the shadow of a point on a jDlane 
is where the ray of light through the point intersects its projec- 
tion on that plane. 

To find the shadow of the line dn on the top of the cube. 
Fig. 67. 

A ray of light through the point n is nn^, its projection on 



124 ISOMETRIC AL DRAWING. 

the top of the cube is dn^ ; these two lines intersect at the point 
72,, which is the shadow of the point n on the top of the cube. 
Join this point with c?, and dn^ is the shadow of the vertical line 
dn required. 

To find the shadow of the line do on the left^ front face of the 
cube. 

A ray of light through the point o is oOg, its projection on the 
face of the cube is do^ ; these lines intersect at the point Og, 
which is the shadow of the point on the left, front face of the 
cube. Join this with d, and do^ is the shadow of the line 
required. 

131. Prob. 37. To find the shadow of a cube on the plane 
of its base. Fig. 67. 

The shadow of the edge ec is ec^ ; of the back edge ac is 
cttg ; of the point b h b^-, therefore, the shadows of ab and he 
are afig and b^Cg respectively, which completes the shadow re- 
quired. 

132. In Figs. 71 and 78 the shadows cast by the objects on 
each other and on the ground are shown. 

133. To find the shadow of any point on any horizontal 
isometric plane proceed as follows : Draw through the point a 
vertical line, and make it equal in length (^downward from the 
point) to the height of the point above the plane receiving the 
shadow ; through the upper end of this line draw a line at 30° 
to a horizontal in the direction of the ray of light ; through the 
lower end draw a horizontal line ; where these two last lines 
intersect will he the shadow of the point. 

To find the shadow of a point on any vertical isometric 
plane : Draw through the point a line at 30° to a horizontal^ 
backward and to the right, and make it equal in length (hack- 



ISOMETEICAL DRAWING. 125 

ward from the point) to the 'perpendicular distance of the point 
from the plane ; through the front end of this line draw a. line 
parallel to the ray of light ; through the hack end draw a line at 
60° to a horizontal, forward and to the right ; where these two 
last lines intersect will he the shadow of the 'point, 

OBLIQUE PROJECTIONS. 

134. Oblique projections differ from isometrical projections 
only in the position of the principal faces of the object. In- 
stead of being placed so that its principal faces make equal 
angles with the plane on which it is represented, one of them 
is placed parallel to this plane, while the edges which represent 
the remaining dimension of the object may be drawn at any 
angle with a horizontal, for convenience usually at 30° or 45°. 
With this difference, all the statements and principles of iso- 
metrical drawing are equally applicable to oblique projections. 

135. Fig. 80 shows the oblique projection of a cube, the 
shadow of lines on its top and front fa€es, the shadow of the 
cube on the plane of its base, and the manner of drawing the 
oblique projection of circles in the three faces of the cube. 
The ellipse in the top face is drawn by the arcs of circles, and 
that in the right face by points located exactly by coordinates. 

It will be noticed that the ray of light is parallel to the diag- 
onal df of the cube, and that its projection on a principal sur- 
face is parallel to the diagonal dh or de of that surface. 

136. Fig. 81 shows the oblique projection of the hollow 
cylinder given as in Fig. 27, of which Fig. 79 is the isometri- 
cal drawing. Figs. 1, 35, and 36 are oblique projections of 
models representing the principles of projections and shadows. 



CHAPTER VII. 



WORKING DRAWINGS. 



137. A working drawing is one which shows all the dimen- 
sions of an object in such a way that the object could be repro- 
duced or constructed from the drawing. Two views at least, 
plan or horizontal projection, and elevation or vertical projec- 
tion, are necessary ; but more frequently a third view, usually 
an end view, is also necessary. Besides these views, one or 
more sections depending upon the object are sometimes neces- 
sary to completely determine all of its dimensions. 

138. If an object is cut through, in any direction, by an 
imaginary plane, the projection of one part of the object on a 
plane parallel to the cutting plane, the person supposed to l)e 
facing this cutting plane, is called a section. Sections are taken 
to show the form and dimensions of the interior of a hollow 
object, and also of some parts of solid objects which are not, 
completely determined by its plan and elevations. These im- 
aginary cutting planes are usually taken either vertical or hori- 
zontal, but it is sometimes necessary to take them in other 
positions, but perpendicular to a vertical or horizontal plane. 
All that part of the object which is cut by these imaginary 
planes is cross-hatched, that is, is covered by parallel lines quite 
near together. The direction of these lines should be either 
30°", 45°, or 60°, and from gV to -J" apart, depending on the size 

126 



Pla,t&7. 




WORKING DRAWINGS. 127 

of the surface to be cross-hatched, the smaller the nearer together. 
They may be drawn in either direction. 

In drawing a section, not only that part of the object which 
is cut by the imaginary plane is represented, but also the pro- 
jection of all parts of the object behind the cutting plane. 

139. Fig. 82 represents the elevation. Fig. 83 the plan, and 
Fig. 84 a vertical section through the centre on the line AB, 
of a stuffing-box gland. The side elevation was not necessary 
in this case, as all the dimensions of the solid are shown with- 
out it. These figures are made one-half size. 

140. It is not sufficient simply to draw the projections of the 
object the correct size, but the dimensions of the solid should 
all be clearly placed on the drawing, so that the workman, or 
whoever has occasion to read the drawing, is not obliged to 
use the scale, thereby removing a great liability to error. These 
dimensions should be put in neatly, and should follow a certain 
system. The method used in Figs. 82, 83, and 84 should be 
carefully observed, besides which the following general direc- 
tions for putting in dimensions and representing special features 
should be followed : — 

In placing dimensions upon the drawing a line should be 
drawn from one point to another, between which the dimension 
is to be given, and the actual dimension, or distance apart of 
the points, is placed in the line, a space having been left for it 
near the centre. These lines should be fine, and composed of 
dashes about -I" long with about ^" spaces between them. When 
the dimension is small of course the dashes must be made 
shorter. Arrow heads are placed at the ends of these lines, 
the point of the arrow exactly touching the points or lines be- 
tween which the dimension is given, the arrow heads pointing 
away from each other. When the dimension is very small the 
arrow heads may be placed on the outside of the lines instead 



128 WORKIi^G DRAWINGS. 

of between tliem, and in that case should point toward each 
other. 

The arrow heads should be drawn free-hand, and not made 
with the drawing pen. The figures for dimensions should also 
be made free-hand, and should always be placed at right angles 
to the dimension line, and should read from the bottom or right- 
hand side of the drawing. They should be put down in inches 
and thirty-seconds, sixteenths, eighths, quarters, and halves, as 
the case may be, thus, 1^^^", not f ^". The fractions should be 
reduced to their lowest terms, thus, f ", not j^f ". The dividing 
line iu the fraction should always be made parallel to the direc- 
tion of the dimension line (see compound fractions, on page 
16, the last way of making the fraction, that is, with the divid- 
ing line of the fraction oblique, unless very carefully made, 
can be easily mistaken for ^^"). The inch marks should be 
placed after the fraction, not between the whole number and 
fraction. 

The whole number should be made about |" high, and the 
total height of the fraction about ^'. 

On rough castings measure to the nearest sixteenth, on ordi- 
nary finished surfaces take the nearest thirty-second, and on 
fine finish and fits be as accurate as possible. 

All dimensions up to twenty-five inches should be put down 
in inches, thus, 15", 22f", and all above that in feet and inches, 
thus, 3-2", or 3 ft. 2 in. Students should be very careful to 
get all of the important dimensions on, and also an " overall " 
dimension, so that the workman will not be required to add a 
number of dimensions together. Important dimensions are 
those which are necessary for the workman to construct the 
piece. The dimensions should not interfere with each other, 
and care should be taken not to have them cross each other in a 
circle. 



WORKING DRAWINGS. 129 

As a general thing do not repeat any dimensions ; that is, if 
a dimension is given on one view do not repeat it on another. 

In order that the drawing may be left as distinct as possible, 
ii is frequently advisable to put the dimensions outside the figure, 
or better, where two or more views are given, put them between 
the different views, as shown in Figs. 82, 83, and 84; the 
widths being placed between the plan and elevation, and the 
heights between the elevation and section, or between the two 
elevations where an end elevation is given. To do this " exten- 
sion " lines must be used. They should be composed of fine 
dash lines, the dashes being about -J" long, so as to be distin- 
guished easily from the dotted lines representing the invisible 
parts of an object. Where the dimensions do not interfere 
with the drawing, as is the case in Fig. 84, it is better to put 
them on the figure between the lines themselves, or as near as 
possible. 

Give the diameter of a circle instead of the radius. When 
only an arc is shown give the radius, and draw a very small 
circle about its centre, and let this circle take the place of an 
arrow head. The dimension line should be drawn from the 
edge of this circle and not from its centre. 

In locating holes or bolts the dimensions should be given 
from the outside of the piece to the centre of the hole or bolt, 
and their distance apart is shown by giving the distance be- 
tween centres. See Fig. 85. Holes are very often located 
from the centre line of a piece, so that it is unnecessary to give 
the dimension from the outside of the piece. 

The " centre " line should be composed of long and short 
dashes, the long dashes being about J" long and the short ones 
about -I" long. 

If the holes are arranged in a circle, as in Fig. 86, give the 
diameter of the circle passing through the centre of the holes. 



130 WORKING DRAWINGS. 

In drawing a bolt or screw represent the threads as shown in 
Fig. 87 ; it is not necessary that the spaces should correspond 
with the true pitch of the threads. In order to obtain the cor- 
rect slant of the threads, a line drawn at right angles to the 
axis of the bolt should pass through the point of a thread on 
one side, and the centre of a space on the other. 

Always give that view of a square, hexagonal, or octagonal 
bolt-head, or nut which shows the distance between its parallel 
sides. 

In placing the dimensions upon a bolt or screw, always give 
the length of the unthreaded part in addition to the length of 
the bolt. The length of the bolt should be given from the 
under side of the head to the extreme end. See Fig. 87. 

In making a sectional view on a line passing lengthwise 
through the centre of a shaft, bolt, or screw, it is generally 
unnecessary to represent the shaft, bolt, or screw in section, as 
the view is more clearly shown by leaving them in full. See 
Fig. 89. 

In drawings where bolts or screws are shown by dotted lines 
do not dot in the threads, but represent them by double dotted 
lines, as shown in Fig. 89. 

Represent a tapped hole as shown in Fig. 90. 

The line on which a section is taken, as AB, Fig. 83, should 
be made the same as a centre line. 

141. Although it is customary to represent a screw by 
straight lines, as shown in Fig. 87, it is sometimes desirable to 
make its actual projections, especially if the screw be a large 
one. 

The thread of a screw is a curve which is called a helix. A 
cylindrical helix is generated by a point caused to travel around 
a cylinder, having, at the same time, a motion in the direction 
of the length of the cylinder, — this longitudinal motion bear- 



WOEKING DRAWINGS. 131 

ing some regular prescribed proportion to the circular or angu- 
lar motion. The distance between any two points which are 
nearest to each other, and in the same straight line parallel to 
the axis of the cylinder, is called the pitch, — in other words, 
the longitudinal distance traversed by the generating point dur- 
ing one revolution. 

To draw the projections of a helix. Fig. 91. 

The plan of the helix will be a circle. Divide this circle 
into any number of equal parts, in this case twelve ; divide the 
pitch into the same number of equal parts. It is evident that 
when the point has moved one-twelfth the distance around the 
circumference, it has also moved in the direction of the axis 
one-twelfth of the pitch ; when it has moved two-twelfths the 
distance around it has moved two-twelfths of the pitch ; there- 
fore, from the points of division, a^, h^, c^, etc., in the plan draw 
vertical lines until they intersect horizontal lines drawn from 
the corresponding division of the pitch. And these intersec- 
tions, a^, If, c", etc., will be points on the vertical projection oi 
the helix. 

Fig. 92 shows a V-threaded screw in projection. 



CHAPTER VIIL 



EXAMPLES. 



All polygons referred to in these Examples are regular polygons, unless other- 
wise stated. 

1. Draw the two projections of a point 1|-" from H and 1" 
from Y. 

2. Of a point lying in H and f " in front of V. 

3. Of a point lying in V and 1" above H. 

4. Of a line 1" long, parallel to both V and H, f" above H 
and 1" in front of V. 

5. Of same line when it is perpendicular to V and 1^" 
above H, its back end being -J" in front of Y. 

6. Of same line when it is perpendicular to H and 1" in 
front of Y, its lower end being ■^" above H. 

7. Of same line when it is parallel to H, 1" above H and 
making an angle of 30° with Y, its back end being ^" in front 
of Y. 

8. Of same line when it is parallel to Y, f " in front of Y 
and making an angle of 60° with H, its lower end being -J" 
above H. 

9. Of same line lying in H and making an angle of 45° 
with Y, its back end being ^" in front of Y. 

10. Of same line lying in Y, parallel to and 1" above H. 

11. Of same line when it is inclined at an angle of 60° with 

132 



Fig.84 



^ 






I" 



i 



■r 



m^' 



j^ 




r 
I 



.88 



Fig. 82 



Plate & 



^f^ Fig.84 ^ 



Fig. 92 




Fig.85 



LJ^^ 





'»■' nail 


V i 


II II 1 -l / 




o ol 



Flg.89 



Fig. 90 



EXAIMPLES. 133 

H, and whose horizontal projection makes an angle of 45° with 
GL, one end being J" above H and 1" in front of V. Line 
slopes downward, forward, and to the left. 

12. Of same line when it makes an angle of 30° with V, 
and whose verticar projection makes an angle of 60° with GL. 
The line slopes downward, backward, and to the left, and passes 
through a point |-" from V and H. 

13. Same as Ex. 12, except that line slopes downward, 
forward, and to the left, and passes through a point ^" from H 
and 1" from V. 

14. A wire l-g-" long projects from a vertical wall at 60° 
with the surface, and is parallel to the ground and 1" above it. 
Draw plan and elevation. 

15. Draw projections of a line 1^" long lying in a profile 
plane and making an angle of 60° with H, sloping downward 
and forward, and passing through a point -J" from V and 1-J-" 
from H. 

16. Of same line lying in a profile plane and making an 
angle of 30° with V, sloping downward and backward, and 
passing through a point ^" from V and H. 

17. Same as Ex. 16, except that line slopes downward and 
forward, and passes through a point |-" from V and 1^" from H. 

18. Draw projections of a line indefinite in length lying in 
a profile plane and making an angle of 75° with H sloping 
downward and forward, and passing through a point -J-" from V 
and ly from H. Locate a point on this line which shall be 1" 
from the fixed point. Locate the point on the line where it 
pierces H. 

19. Find true length of a line given by its projections, as 
follows : One end is ^" from each plane and the other is 2" 
above H, the horizontal projection of the line is 1^" long and 
makes an angle of 30° with GL. 



134 EXAMPLES. 

20. Draw plan and elevation of an oblique line, one end 
being above and in front of the other. Find its true length and 
angle it makes with H. 

21. Of an oblique line, one end bekig behind and above the 
other. Find its true leno^th and ano-le it makes with V. 

22. Of a line which slopes downward, backward, and to the 
right. Find its true length by revolving parallel to V. 

23. Of a line which slopes downward, forward, and to the 
left. Find its true length by revolving parallel to H. 

24. Of two lines which are parallel in space and slope 
downward, forward, and to the right. 

25. Of a line 2" long, sloping downward, forward, and to 
the right, one end being 1-J" above H and I-" in front of V, the 
other end ^" above H and IJ" in front of Y. 

26. Draw plan and elevation of a rectangular card f" x 1^" 
which is perpendicular to H, parallel to V, and f" in front of 
V ; its short sides are parallel to H and the lower one is |-" 
above H. E-evolve this card forward about its left-hand edge 
(like a door on its hinges) through angles of 30°, 45°, 60°, and 
90°, and construct corresponding plans and elevations. 

27. Of same card when it is lying on H with its long sides 
parallel to V and J" in front of V. Revolve card about right- 
hand horizontal edge (like a trap-door on its hinges) through 
angles of 30°, 45°, 60°, and 90°, and construct corresponding 
projections. 

28. Of same card when, besides making the angles of 0°, 
30°, 45°, 60°, and 90° with H, as in last example, the hori- 
zontal projection of its long edges, in all its different posi- 
tions, makes an angle of 30° with GL backward and to the 
right. 

29. Of same card when it is parallel to V and ^" in front 
of V, one of its diagonals being parallel to H. Revolve this 



EXAMPLES. 135 

card through an angle of 60° about a vertical axis, and construct 
its corresponding projections. 

30. Of same card when, besides making an angle of 60° 
with Y, as in last example, the vertical projection of the diago- 
nal which was parallel to H makes an angle of 45° with GL. 

31. Of a card |-" square resting on H, with one diagonal 
parallel to V and 1" in front of V, then raise left-hand end of 
diagonal until it makes an ano-le of 45° with H and construct 
corresponding projections. 

32. Of the same card when, besides making an angle of 45° 
with H, as in last example, the horizontal projection of the 
diaofonal makes an ano-le of 30° with GL. 

33. Same as example 32, except that the horizontal projec- 
tion of the diawnal makes an angle of 90° with GL. 

34. Of an hexagonal card whose sides are |-" long, when 
one of its long diameters is parallel to Y and 1" in front of Y, 
one end resting on H. The surface of the card is perpendicu- 
lar to Y and makes an angle of 30° with H. 

35. Of same card when its diameter, besides making an 
angle of 30° with H, as in last example, has its horizontal pro- 
jection inclined at an angle of 60° with GL, and slopes down- 
ward, backward, and to the left. 

36. Of a pentagonal card whose surface is perpendicular to 
H and makes an angle of 45° with Y, its left-hand edge being 
perpendicular to H and resting against Y. The diameter of 
the circumscribed circle is 1|-". 

37. Of same card when, besides making an angle of 45° 
with Y, as in last example, it has been revolved through an 
angle of 30°. 

38. Of an octagonal card resting on one of its edges, wdth 
its surface perpendicular to Y and inclined at an angle of 60° 
with H. The diameter of the inscribed circle is 1|-". 



136 EXAMPLES. 

39. Of same card when it is inclined at same angle with H 
as in last example, and the horizontal projection of the edges 
which were perpendicular to Y make angles of 45° with GL. 

40. Draw the projections of a card 1|" square when its 
surface is perpendicular to V and makes an angle of 30° with 
H, the H projection of the diagonal making an angle of 30° 
with GL, one end of the diagonal resting on H. 

Revolve plan last found through an angle of 30° and con- 
struct corresponding elevation. 

41. Of an isosceles triangle situated in a profile plane, the 
base making an angle of 15° with H, its back corner being f" 
above H and f" in front of V, and lower than its front corner. 
The base of triangle is IJ" and altitude 1". 

42. Of an hexagonal card whose surface is perpendicular to 
both V and H. Its long diameter is 1". Two of its edges are 
perpendicular to H. Centre of card is 1" above H, and 1^" in 
front of V. 

43. Of a card f " square lying in a profile plane. Its lower 
front edge slopes downward and backward, and makes an angle 
of 30° with H. The lowest corner of card is |-" from H, and 
f" from V. 

44. Of a circular card 3" in diameter, perpendicular to H 
and making an angle of 30° with Y, resting on H, back edge 
1" from Y. 

45. Of same card after it has been revolved through an 
angle of 45°, keeping the same angle with Y as in last example. 

46. Of a circular card 3" in diameter, perpendicular to H 
and making an angle of 45° with Y. Solve as explained in 
Art. 60. 

Shade lines are to be put in in all the examples where there are any. 

47. Draw plan and elevation of a cube of 1" edge resting 
on H, i" in front of Y, with two faces parallel to Y. Find 



EXAMPLES. 137 . 

the true size of the angle which the diagonal, which slopes 
downward, backward, and to the right, makes with V and also 
with H. 

48. "* Of a rectangular prism whose base is f" x 1" and 
length is 1^" resting with its base against Y, its lower left-hand 
face making an angle of 60° with H. 

49. Of a cylinder resting with its base on H ; diameter of 
cylinder is 1" and its length is IJ". The axis is f" in front 
of V. 

50. Of a cone resting with its base parallel to and J" in 
front of V. The diameter of base is 1^" and the height of 
cone is If". 

' 51. Of a heptagonal prism resting with its base on H, one 
of its faces making an angle of 15° with V. The diameter of 
the circumscribed circle about base is 1" and the height of 
prism is If". 

52. Of an octagonal pyramid resting with its base on H, 
with two of the edges of the base making an angle of 30° with 
GL. The diameter of the circle inscribed in the base is 1-J-" 
and the altitude is |-". 

53. Draw plan and two elevations of a square prism with 
its axis parallel to both V and H ; the axis is 1" above H and 
1^" in front of Y, base of prism is f " square, and the length is 
1^"; the upper front long face makes an angle of 30° with H. 

54. Draw plan and elevation of same prism with its surfaces 
making same angle with H, when its axis is parallel to H and 
makes an angle of 45° with Y, backward and to the left. 

55. Draw the plan and elevation of a regular triangular 
prism 2" long, with its axis parallel to H, but making an angle 
of 60° with Y, backward and to the left. The axis is 1" above 

* The base is supposed to be at right angles to the axis in all the prisms, pyra^ 
mids, cylinders, and cones, unless otherwise stated. 



138 EXAMPLES. 

H, and its front end is 2J" in front of V. The lower right- 
hand long face of prism makes an angle of 15° with H. Diam- 
eter of circle circumscribed about base is 2". 

56. Draw plan and two elevations of an hexagonal prism 
2" long, diameter of circumscribed circle about base is 1^", the 
axis is parallel to H and makes an angle of 30° with V, back- 
ward and to the left ; lower edge of prism rests on H, and the 
lower ri^ht-hand face makes an angle of 10° with H. 

57. Draw plan and two elevations of a circular cylinder, 
3" in diameter and 2^" long, with a circular hole through it 
1^" in diameter ; axis is parallel to both V and H. 

58. Draw plan and elevation of same cylinder resting with 
its base on H. 

59. Of same cylinder when lying on H, with its axis par- 
allel to H and making an angle of 60° with Y. 

60. Draw plan and two elevations of a pile of blocks 
located as follows : the lowest one is 3" long, 1^" wide, and ^" 
thick, it rests with its wide face on H, its long edge making an 
angle of 30° with V, backward and to the right ; on top of this 
a second block rests, equal in width and thickness to the first 
but 1" shorter ; on this a third block rests, of the same width 
and thickness as the others, but 1" shorter than the second; 
these blocks are placed symmetrically. 

61. Draw plan and elevation of a pyramid formed of four 
equilateral triangles of 2" sides, when one edge of the base is 
at 45° with V. 

62. Of an hexagonal prism standing with its base on H, 
two of its faces making angles of 20° with V, backward and to 
the left ; diameter of circumscribed circle about base is 8", 
length of prism is 10". Scale 3" = 1', or J size. 

63. Of same prism when axis is parallel to V and makes 
an angle of 60° with H and slopes downward to the left. 
Scale 3"= 1'. 



EXAMPLES. 139 

64. Of same prism when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an angle 
of 60° with GL ; axis of prism slopes downward, forward, and 
to the left. Scale 3" =1'. 

65. Of a pentagonal pyramid resting with its base on H, 
left-hand edge of base perpendicular to V, diameter of circum- 
scribed circle 1 6", height of pyramid 20". Find its shadow. 
Scale l|-" = r, or -J size. 

66. Of same pyramid when its axis is parallel to V and 
slopes downward to the left, making an angle of 75° with H. 
Find its shadow. Scale 1|-" = 1'. 

67. Of same pyramid when its axis, besides making an 
angle of 75° with H, has its horizontal projection inclined at 
an angle of 30° with GL, so that axis slopes downward, back- 
ward, and to the right. Find its shadow. Scale l|-"=r. 

68. Of a cone resting with its base on H, diameter of base 
20", height of cone 2'. Find its shadow. Scale H"=l'. 

69. Of same cone when resting with an element on H, with 
its axis parallel to V. Find its shadow. Scale 1^" = 1'. 

70. Of same cone with an element on H, and axis making 
an angle of 45° with GL. Find its shadow. Scale 1J"=1'. 

71. Of same cone with an element on H, and axis lying 
in a profile plane, sloping downward and backward. Find its 
shadow. Scale 1^" = 1'. 

72. Of the frustum of an octagonal pyramid resting with 
its base on H, long diameter of lower base is 3" and of the 
upper base is 2", the height of frustum is 3", the front left-hand 
edge of base makes an angle of 15° with GL, backward to the 
left. There is a hole 1" square through the centre of frustum, 
whose axis is coincident with axis of frustum ; two sides of the 
hole make angles of 7|-° with GL. 

73. Of same frustum when its axis is parallel to V and 
makes an angle of 60° with H and slopes downward to the left. 



140 EXAMPLES. 

74. Of same frustum when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an angle 
of 30° with GL, and slopes downward, forward, and to the left. 

75. Of the skeleton frame of a box 3' long, 2' wide, and 2' 
high, the joists being all 3" square. The frame rests on H 
with its long sides parallel to Y. Do not show joints in fram- 
ing. Scale 1" = 1'. 

76. Of same frame still restinoj on H with its Ions; sides mak- 
ino; an angle of 50° with V, backward to the left. Scale 1" = 1'. 

77. Revolve the elevation obtained in example 76 through 
an angle of 30° (in either direction), and construct correspond- 
ing plan. Scale 1"=:1'. 

78. Of a double cross standing on its base, one arm paral- 
lel to both V and H ; upright piece is l'-8" square and 10'-8" 
high, each arm is l'-8" square and 7' long (out to out), their 
top surfaces are 2-8" below the top of upright. Scale f" = l'. 

79. Of same cross when its axis is parallel to V and makes 
an angle of 60° with H. Scale f'^rl'. 

80. Of same cross when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an 
angle of 60° with GL, sloping downward, forward, and to the 
left. Scale f" = l'. 

81. Of a pyramid resting on its apex with axis perpendicu- 
lar to H and 2'-6" in front of V, its base is an equilateral tri- 
au2;le of 2'-8" sides and its altitude is 11". The left-hand edge 
of base is perpendicular to Y. Find its shadow. Scale f" = l'. 

82. Of same pyramid when its axis is parallel to Y and 
makes an angle of 60° with H, and slopes downward to the 
right. Find its shadow. Scale f" = l'. 

83. Of same pyramid when its axis, besides making an 
angle of 60° with H, has its horizontal projection inclined at 
an angle of 30° with GL, so that it slopes downward, forward, 



EXAMPLES. 141 

and to the left, its apex being 2-6" from V, still resting on H. 
Find its shadow. Scale f "= 1'. 

Draw projections of a line indefinite in length, sloping down- 
ward, forward, and to the right, through a point 2' in front, and 
6" to the right, of apex of pyramid. Line makes an angle of 
75° with V, and its Y projection makes an angle of 60° with 
GL. Same scale. 

Find its shadow on H and V, and also on pyramid. 

84. There is a solid formed of two equal square pyramids 
of 2" base and 3" altitude, which are united by their bases. 
Draw plan and elevation when the object rests with one of its 
triangular faces on H, its axis being parallel to Y and 2|-" in 
front of Y. Find its shadow. 

85. Of same object still resting on one of its faces, when 
the horizontal projection of the axis makes an angle of 45° 
with GL, and slopes downward, backward, and to the right. 
Find its shadow. 

86. Draw the plan and elevation of the frustum of a square 
pyramid resting on its base, with two of the edges of the base 
perpendicular to Y. Lower base is 3" square, upper base is 
1" square, and the altitude is 2^". 

In the centre of each of the sloping faces there is a square 
of 1" side, so placed that one of its diagonals is parallel to H. 

87. Draw projections of a pentagonal prism whose length 
is 2^" and radius of circumscribed circle about the end is -J"; 
the prism rests with one of its long edges on H, which makes 
an angle of 60° with Y, backward to the left, and whose front 
end is 3^" from Y. The lower left-hand long face makes an 
angle of 15° with H. 

Also, of a triangular pyramid resting on its base on H, with 
its axis 2" to the left of the point located in prism, and 4^" in 
front of Y, diameter of circumscribed circle is 2", the altitude 



142 EXAMPLES. 

of pyramid is 3f", right-hand edge of base is perpendicular to 
V. Find shadow of prism on H and V, also of pyramid on H 
and on prism. 

88. Draw projections of a prism whose base is 1^" square 
and length is 2^". Prism rests on H on one of its long edges, 
which makes an angle of 60° with V, backward and to the 
right, and whose front end (of edge) is 3" from V. Lower 
left-hand long face makes an angle of 30° with H. 

Also, of an indefinite line sloping downward, forward, and to 
the left, passing through upper corner of the front end of 
prism. Line makes angle of 75° with H, and its H projec- 
tion makes 80° with GL. 

Find shadow of prism on H and V, also of line on H and 
Y and on the prism. 

89. Draw the projections of a solid as follows : A frustum 
of a square pyramid rests with its base on H, two of the edges 
of the base make angles of 30° with Y, backward to the right. 
Lower base is 2" square, upper base is If" square, and its height 
is 2^". Axis of frustum is 2^" from Y. 

On top of this frustum is a square pyramid, whose base is 
the same size as the top of frustum and coincides with it. 
Height of pyramid is 1". Find shadow of solitl on Y and H. 

Draw, projections of an indefinite line, sloping downward, 
forward, and to the right, passing through upper front corner 
of frustum, making an angle of 30° with Y, and whose Y pro- 
jection makes an angle of 60° with GL. Find shadow of line 
on Y and H, and on the frustum* and pyramid. 

90. Find the shadow on H of a card |" square, parallel to 
H, and ^" above H, two edges making angles of 30° with Y. 
Back corner of card is 1" from Y. 

9L Of same card on H, when it is parallel to Y, 2" in front 
of Y, two edges parallel to H, lowest edge ^" above H. 



EXAMPLES. 143 

92. Of same card on V and H, lying in a profile plane, two 
edges perpendicular to H, back edge -j" in front of V and low- 
est edge ^" above H. 

93. Of an hexagonal card, parallel to Y, ^" in front of Y, 
two edges parallel to H, centre of hexagon 1^" above H, long 
dianaeter 1". 

94. Of same card parallel to H, 1-J" above H, centre 1" in 
front of Y, two edges parallel to Y. 

95. Of a circular card, IJ" in diameter, parallel to Y, 1^" 
in front of Y, centre 1" above H. 

96. Of an hexagonal card whose surface is perpendicular 
to Y and H, two of its edges perpendicular to H, centre of 
hexagon i|" above H, 1-^q" in front of Y, diameter of inscribed 
circle 1". In constructing projections of hexagon revolve it 
about a vertical axis through centre. 

97. Of a cube of f" sides, parallel to Y and H, 1-J" above 
H, and ^" in front of Y. 

98. Of a square prism standing on H, each face -f" x 1|-", 
two faces parallel to Y, ^" in front of Y. 

99. Of same prism still standing on H, turned so that two 
faces make angles of 30° with Y, backward to the right, ^" 
from Y. 

100. Of a cylinder f" in diameter and If" high, with base 
resting against Y, and axis 1" above H. 

101. Of a cone 1:^" high, base f" in diameter, standing on 
H, with axis 1^" in front of Y. 

102. Of a line located as in example 18. 

103. Of a line located as in example 25. 

104. Of a card located as in example 41. 

105. Of a card located as in' example 42. 

106. Of a card located as in example 43. 



144 EXAMPLES. 

107. Fig. 93 shows plan and elevation of two rectangular 
sticks, one on the other, resting on an abutment. Find the 
shadow of the top stick on the top and front faces of the bot- 
tom stick (Art. 103), also of the two sticks on the top and front 
faces of the abutment (Art. 101). 

108. Fig. 94 shows plan and elevation of two sticks framed 
together. Find shadow of the oblique stick on the top of the 
horizontal one (Art. 102). 

109. Change the head in Fig. 57 from octagonal to cylin- 
drical and make it 2" in diameter and ^" thick, make long 
diameter of hexagonal shank 1^", and length of shank 1^". 
Find shadow of head on shank (Art. 105). 

110. Fig. 95 shows two elevations of a roof and chimney. 
Find shadow of chimney on roof (Art. 106). 

111. Fig. 96 shows plan and elevation of a square stick rest- 
ing on top of a hollow semi-cylinder. Find shadow of semi- 
c\ linder on itself, also of the stick on the top and inside of cylin- 
der (Art. 104). 

112. Find shadow of wing wall on flight of steps as shown 
in Fig. 97. 

113. Fig. 98 shows two views of a round stick lying on a 
molding. Find shadow of molding on the ground and on itself, 
also of the stick on molding and on the ground. 

114. Fig. 99 shows the plan of a pile of blocks of which 
the lowest one is prismatic and f" high, the middle one is the 
frustum of a pyramid and is f" high, the top one is a pyramid 
and is 2^" high. Find the shadow of the lowest one on the 
ground ; of the middle one on the top of the lowest and on the 
ground ; also of the top one on the top of each of the under 
ones and on the ground. 

115. Fig. 100 shows plan and elevation of 4 sticks framed 
together. Find shadow of the group on itself and on the ground. 



EXAMPLES. 145 

116. Fig. 101 shows plan and elevation of a fluted column. 
Find shadow on itself. 

Through point marked with a circle draw an indefinite line 
which makes an angle of 15° with H, and whose H projection 
makes 30° with GL. The line slopes downward, forward, and 
to the right. Find shadow of line on top and front of column. 

117. Fig. 102 shows plan and elevation of a rectangular 
stick resting on a pier. Find shadow of stick on pier (Art. 107). 

118. Make the isometric drawing and cast the shadows of a 
cube of If" edge with a prism f " square and ^" long cut out of 
each corner, the length of prism in each case being vertical. 

119. Make the isometric drawing of a box f" high by 1^" 
wide by 2^' long; the thickness of sides, bottom and cover is •^^". 
Let the cover be opened through an angle of 150°. 

120. Make the isometric drawing of the group of blocks 
framed together as shown in Fig. 103. 

121. Make the isometric drawing of the upright block in 
Fig. 103 with the gains cut entirely from the upright. 

122. Fig. 105 shows the plan of the skeleton frame of the 
frustum of a pyramid. The height of the frustum is 2", and 
the vertical thickness of the top and bottom pieces is |-". Make 
the isometric drawing of frustum. 

123. Make isometric drawing of the square, circle, and 
ellipse as shown in Fig. 107 by the exact method. 

124. Make the isometric drawing of a cube of If" edge 
with a cone of If" base and If" altitude projecting from each 
face. Make circles by the approximate method. 

125. Make the isometric drawing of a block 1^" wide by 2" 
long by ^" thick, with round corners of ^" radius. Make cor- 
ners by approximate method. 

126. Make the isometric drawing of group of blocks referred 
to in Ex. 114 as located in Fig, 99. 



146 EXAMPLES. 

127. Fig. 104 shows plan of a carpenter's saw-horse. The 
height of horse is 18" and the thickness of top piece is 3". Make 
the isometric drawing of this horse. Scale l"= 1'. 

128. Make the isometric drawing of double cross shown in 
Fig. 106. Plan and elevation are the same. 

129. Make the isometric drawing of the group as given in 
Ex. 89. 

130. Draw projections of a line 2" long, one end being on 
H i" from V. The H projection of line is 1" long, and makes 
45° with GL. 

131. Draw projections of a line 1" long, when it makes 30° 
with H, and 45° with V, and slopes downward, backward, and 
to the left. Lower end of line is J" from V and H. 

132. Draw projections of a line 2" long, when it makes 45° 
with H, and 30° with V, and slopes downward, forward, and to 
the right. Upper end is ^" from V, and If" from H. 

133. Draw projections of a pyramid, whose base is 2" square, 
and whose slant height is 3"; base is on H. Left-hand corner 
is 2" from V; left-hand front edge of base makes 30° with V, 
forward, and to the right. 

134. Draw plan and elevation of the frustum of a square 
pyramid resting on its base, with two of the edges of the base 
30° with V, backward and to the right. Lower base is 3" square, 
upper base is 1" square, and its sloping edges are 2^". 

In the centre of each of the sloping faces there is a square of 
1" side, so placed that one of its diagonals makes 30° with edge 
of base of frustum. Back corner is §-" from V. 



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